KCET 2020 Physics Questions with Answers Key Solutions

Q1. The value of acceleration due to gravity at a height of $10\,km$ from the surface of earth is $x$ . At what depth inside the earth is the value of the acceleration due to gravity has the same value $x$ ?

A

5 km

B

20 km

C

10 km

D

15 km

Solution:

$g_{h}=g\left(1-\frac{2 h}{R}\right)$
$g_{d}=g\left(1-\frac{d}{R}\right)$
$\Rightarrow g_{h}=g_{d}$
$g\left(1-\frac{2 h}{R}\right)=g\left(1-\frac{d}{R}\right)$
$\Rightarrow d=2 h=2 \times 10=20\, km$

Q2. Young’s modulus of a perfect rigid body is

A

Zero

B

Unity

C

Infinity

D

Between zero and unity

Solution:

Young’s modulus = stress/strain, and for a perfect rigid body we know that strain is equal to $0$ and hence Young’s modulus of a perfect rigid body becomes Infinity.

Q3. A wheel starting from rest gains an angular velocity of $10$ rad/s after uniformly accelerated for $5$ sec. The total angle through which it has turned is

A

25 rad

B

100 rad

C

25 $\pi$ rad

D

50 $\pi$ rad about a vertical axis

Solution:

$\omega_{1}=0$
$\omega_{2}=10 \frac{ rad }{ sec } t=5 \,sec$
$\theta=\frac{\omega_{1}+\omega_{2}}{2} \times t$
$=\frac{10}{2} \times 5=25 \,rad$

Q4. Iceberg floats in water with part of it submerged. What is the fraction of the volume of iceberg submerged if the density of ice is
$\rho_i = 0.917 \,g\,cm^{-3}$ ?

A

0.917

B

1

C

0.458

D

0

Solution:

Let $V$ be the total volume of the iceberg and $V'$ of its volume be submerged into water
Floatation condition weight of iceberg = Weight of water displaced by submerged part by ice
$V \rho_{i} g=V' \rho_{w} g$
$\Rightarrow V' / V=\rho_{i} / \rho_{w}$
$=0.917 / 1=0.917$

Q5. A sphere, a cube and a thin circular plate all of same material and same mass initially heated to same high temperature are allowed to cool down under similar conditions. Then the

A

plate will cool the fastest and cube the slowest

B

sphere will cool the fastest and cube the slowest

C

plate will cool the fastest and sphere the slowest

D

cube will cool the fastest and plate the slowest

Solution:

Surface area is more for plate and less for sphere. Hence plate will cool the fastest and sphere the slowest.

Q6. In an adiabatic expansion of an ideal gas the product of pressure and volume

A

Decreases

B

Increases

C

Remains constant

D

At first increases and then decreases

Solution:

In an adiabatic expansion as temperature decreases from an ideal gas equation $PV=nRT$ .
Product of pressure and volume also decreases.

Q7. A certain amount of heat energy is supplied to a monoatomic ideal gas which expands at constant pressure. What fraction of the heat energy is converted into work ?

A

1

B

$\frac {2}{3}$

C

$\frac {2}{5}$

D

$\frac {5}{7}$

Solution:

The heat $Q$ is converted into the internal energy and work. According to the first law of thermodynamics,
$Q = U + W$
$\Rightarrow nC _{ p } \Delta t = n C _{ v } \Delta t + W$
$\frac{W}{Q}=\frac{n C_{p} \Delta t-n C_{v} \Delta t}{n C_{p} \Delta t}$
$\frac{c_{p}-c_{v}}{c_{p}}=1-\frac{1}{\gamma}$
$=1-\frac{1}{\frac{5}{3}}=\frac{2}{5}$

Q8. A tray of mass $12 \,kg$ is supported by two identical springs as shown in figure. When the tray is pressed down slightly and then released, it executes SHM with a time period of $1.5\,s$ . The spring constant of each spring is

A

$50 \,Nm^{-1}$

B

$0$

C

$105 \,Nm^{- 1}$

D

$\infty$

Solution:

We know that
$T=2 \pi \sqrt{\frac{m}{k_{\text{eff}}}}$
$\frac{3}{2}=2 \pi \sqrt{\frac{12}{2 k}}$
$\Rightarrow \frac{9}{4}=4 \pi^{2} \times \frac{12}{2 k} k \approx 105 \,Nm ^{-1}$

Q9. A train whistling at constant frequency ‘ $n$ ’ is moving towards a station at a constant speed $V$ . The train goes past a stationary observer on the station. The frequency ‘ $n$ ’ of the sound as heard by the observer is plotted as a function of time ‘ $t$ ’. Identify the correct curve

A

B

C

D

Solution:

Whistling train is the source of sound, $v_{s}=B$ . Before crossing a stationary observer on station, frequency heard is $n^{\prime}=\frac{v n}{\left(v-v_{s}\right)}=\frac{v n}{v-V}=$ constant, and $n^{\prime}>n^{2}$ . Here , $v$ is velocity of sound in air and $n$ is actual frequency of whistle. After crossing the stationary observer, frequency heard is $n^{\prime}=\frac{v}{\left(v+v_{s}\right)}=\frac{v}{v+V}=$ constant and $n^{\prime}

Q10. A point charge ‘ $q$ ’ is placed at the corner of a cube of side ‘ $a$ ’ as shown in the figure. What is the electric flux through the face $ABCD$ ?

A

$0$

B

$\frac {q}{24\epsilon_o}$

C

$\frac {q}{6\epsilon_o}$

D

$\frac {q}{72\epsilon_o}$

Solution:

ACCORDING TO THE GIVEN PROBLEM, THE CHARGE IS PLACED AT THE CORNER OF THE CUBE AS SHOWN IN FIGHURE_1,

NOW CONSIDER THE ANOTHER FIGURE_2 IN WHICH THE WHOLE CHARGE IS ENCLOSED IN THE FOUR SUCH CUBE,

NOW CALCULATING THE FLUX THROUGH THE ALL CUBES AS TOTAL:
$\phi_{ e }=\frac{ q }{\epsilon_{0}}= EA$
WHERE A $=6 *(2 a)^{2}$
FOR ONLY ONE SIDE OF FIRST CUBE,
AREA WILL BE A ${ }_{0}= a ^{2}$
FROM ABOVE DATA FLUX THROW THE SINGLE FACE (AREA = $A _{0}$ ) IS GIVEN BY:
$\phi_{ e }=\frac{ Q }{\epsilon_{0}}= EA _{0}$
WHERE Q $=\frac{q}{24}$
SO THE FLUX THROUGH THROUGH THE SINGLE FACE IS $\frac{ q }{24 \epsilon_{0}}$ ;

Q11. The electric field lines on the left have twice the separation on those on the right as shown in figure. If the magnitude of the field at $A$ is $40\,Vm^{-1}$ , what is the force on $20\,\mu \,C$ charge kept at $B$ ?

A

$4 \times 10^{-4} Vm^{-1}$

B

$8 \times 10^{-4} Vm^{-1}$

C

$16 \times 10^{-4} Vm^{-1}$

D

$1 \times 10^{-4} Vm^{-1}$

Solution:

We know that
$F = qE =20 \times 20 \times 10^{-6}$
$=4 \times 10^{-4} V / m^{-1}$

Q12. An infinitely long thin straight wire has uniform charge density of $\frac {1}{4} \times 10^{-2} cm^{-1}.$ What is the magnitude of electric field at a distance $20\, cm$ from the axis of the wire ?

A

$1.12 \times 10^8 \,NC^{-1}$

B

$4.5 \times 10^8 \,NC^{-1}$

C

$2.25 \times 10^8 \,NC^{-1}$

D

$9 \times 10^8 \,NC^{-1}$

Solution:

We know that
$E=\frac{\lambda}{2 \pi \varepsilon_{0} r}$
$=\frac{1}{4} \times \frac{10^{-2}}{10^{-2}} \times 18 \times 10^{9} \times 5$
$=2.25 \times 10^{8} N / C$

Q13. A dipole moment ‘ $P$ ’ and moment of inertia $I$ is placed in a uniform electric field $\vec{E}$ . If it is displaced slightly from its stable equilibrium position, the period of oscillation of dipole is

A

$\sqrt{\frac{PE}{I}}$

B

$2\pi\sqrt{\frac{I}{PE}}$

C

$\frac {1}{2\pi} \sqrt{\frac{PE}{I}}$

D

$\pi \sqrt{\frac{I}{PE}}$

Solution:

Dipole moment $= p$ electric field $= E$ centroid axis $= I$ Explanation When displaced at an angle $\theta$ from its mean position the magnitude of restoring torque is $T =$ $- p \sin \theta$ For small angular displacement $\sin \theta \approx \theta$ $\begin{array}{l} T =- pE \theta \\ \alpha=\frac{ T }{ I }=-\left(\frac{ PE }{ I }\right) \theta \\ =- w ^2 \theta \\ w^2=\frac{P E}{I} \\ T =2 \pi \sqrt{\frac{ I }{ PE }} \\ \end{array}$ (P.E $=$ moment in electric field)

Q14. The difference between equivalent capacitances of two identical capacitors connected in parallel to that in series is $6 \,\mu F$ . The value of capacitance of each capacitor is

A

$2 \mu F$

B

$3 \mu F$

C

$4 \mu F$

D

$6 \mu F$

Solution:

$C _{ P - C }=6\, \mu F$
$2 C -( C / 2)=6$
$C =4\, \mu F$

Q15. Figure shows three points $A, B$ and $C$ in a region of uniform electric field $\vec{E}$ . The line AB is perpendicular and $BC$ is parallel to the field lines. Then which of the following holds good ? $(V_A, V_B$ and $V_C$ represent the electric potential at points $A, B$ and $C$ respectively)

A

$V_A = V_B = V_C$

B

$V_A = V_B > V_C$

C

$V_A = V_B < V_C$

D

$V_A > V_B = V_C$

Solution:

Electric lines of force in an electric field always flow from a higher potential to a lower potential.
Hence, $V _{ A }= V _{ B } > V _{ C }$

Q16. When a soap bubble is charged ?

A

Its radius increases

B

Its radius decreases

C

The radius remains the same

D

Its radius may increase or decrease

Solution:

When the radius of a soap bubble increases, the soap bubble gets charged.

Q17. A hot filament liberates an electron with zero initial velocity. The anode potential is $1200\,V$ . The speed of the electron when it strikes the anode is

A

$1.5 \times 10^5 \,ms^{-1}$

B

$2.5 \times 10^6 \,ms^{-1}$

C

$2.1 \times 10^7 \,ms^{-1}$

D

$2.5 \times 10^8 \,ms^{-1}$

Solution:

$\frac{1}{2} m v^{2}=V q$
$v=\sqrt{\frac{2 V q}{m}}$
$=2.1 \times 10^{7} ms ^{-1}$

Q18. A metal rod of length $10 \,cm$ and a rectangular cross - section of $1 \,cm \times \frac{1}{2}\, cm$ is connected to a battery across opposite faces. The resistance will be

A

maximum when the battery is connected across $1 \,cm \times \frac {1}{2}$ cm faces

B

maximum when the battery is connected across $10 \,cm\, \times \frac {1}{2} cm$ faces

C

maximum when the battery is connected across $10 \,cm \times$ 1 cm faces

D

same irrespective of the three faces

Solution:

Resistance is inversely proportional to area.
Hence resistance will be maximum when the battery is connected across $1 \,cm \times$ $(1 / 2) cm$ faces.

Q19. A car has a fresh storage battery of e.m.f $12\,V$ and internal resistance $2 \times 10^{-2} \Omega$ . If the starter motor draws a current of $80\,A$ . Then the terminal voltage when the starter is on is

A

12V

B

8.4V

C

10.4V

D

9.3V

Solution:

$V=E-ir=10.4 \,V$

Q20. A potentiometer has a uniform wire of length $5\,m$ . A battery of emf $10\,V$ and negligible internal resistance is connected between its ends. A secondary cell connected to the circuit gives balancing length at $200 \,cm$ . The emf of the secondary cell is

A

4V

B

6V

C

2V

D

8V

Solution:

Emf of cell in the secondary circuit =potential gradient $x$ balancing length
$\Rightarrow (10 / 5) \times 2=4\, V$

Q21. The colour code for a carbon resistor of resistance $0.28\,k\Omega \pm 10$ % is

A

Red, Grey, Brown, Silver

B

Red, Green, Brown, Silver

C

Red, Grey, Silver, Silver

D

Red, Green, Silver

Solution:

Red, Grey, Brown, Silver

Q22. Each resistance in the given cubical network has resistance of $1\,\Omega$ and equivalent resistance between $A$ and $B$ is

A

$\frac {5} {6} \Omega$

B

$\frac {6} {5} \Omega$

C

$\frac {5} {12} \Omega$

D

$\frac {12} {5} \Omega$

Solution:

$5 / 6\, \Omega$

Q23. $I-V$ characteristic of a copper wire of length $L$ and area of cross-section $A$ is shown in figure. The slope of the curve becomes

A

More if experiment is performed at higher temperature

B

More if a wire of steel of same dimension is used

C

Less if the area of the wire is increased

D

Less if the length of the wire is increased

Solution:

Slope $=1 / R = A / \rho l$
Hence the slope becomes less if the length of the wire is increased.

Q24. In the given figure, the magnetic field at ‘ $O$ ’.

A

$\frac {3}{4} \frac {\mu_oI}{r}+\frac {\mu_oI}{4\pi r}$

B

$\frac {3}{10} \frac {\mu_oI}{r} - \frac {\mu_oI}{4\pi r}$

C

$\frac {3}{8} \frac {\mu_oI}{r} + \frac {\mu_oI}{4\pi r}$

D

$\frac {3}{8} \frac {\mu_oI}{r} - \frac {\mu_oI}{4\pi r}$

Solution:

$B_{\text {net }}=B_{1}+B_{2}+B_{3}$
$=\frac{3}{8} \frac{\mu_{0} I}{r}+\frac{\mu_{0} I}{4 \pi r}+0$

Q25. The magnetic field at the origin due to a current element $\overrightarrow{id1}$ placed at a point with vector position $\overrightarrow{r}$ is

A

$\frac{\mu_{0}i}{4\pi} \frac{\overrightarrow{d1} \times \overrightarrow{r}}{r^{3}}$

B

$\frac{\mu_{0}i}{4\pi} \frac{\overrightarrow{r} \times \overrightarrow{d1}}{r^{3}}$

C

$\frac{\mu_{0}i}{4\pi} \frac{\overrightarrow{d1} \times \overrightarrow{r}}{r^{2}}$

D

$\frac{\mu_{0}i}{4\pi} \frac{\overrightarrow{r} \times \overrightarrow{d1}}{r^{2}}$

Solution:

$\frac{\mu_{0} i}{4 \pi} \frac{\overrightarrow{d l} \times \vec{r}}{r^{3}}$

Q26. A long cylindrical wire of radius $R$ carries a uniform current $I$ flowing through it. The variation of magnetic field with distance ‘ $r$ ’ from the axis of the wire is shown by

A

B

C

D

Solution:

The magnetic field increases as the point moves closer to the boundary of the wire and decreases as it moves away from the boundary of the wire.

Q27. A cyclotron is used to accelerate protons $\binom{1 H}{1}$ , Deuterons $\binom{2 H}{1}$ and $\alpha$ -particles $\binom{4 He}{2}$ . While exiting under similar conditions, the minimum K.E. is gained by

A

$\alpha$ -particle

B

Proton

C

Deuteron

D

Same for all

Solution:

K.E. $=q^{2} / m$
Hence, a deuteron gains the least kinetic energy.

Q28. A paramagnetic sample shows a net magnetization of $8\, Am^{-1}$ when placed in an external magnetic field of $0.6\,T$ at a temperature of $4\,K$ . When the same sample is placed in an external magnetic field of $0.2 \,T$ at a temperature of $16\, K$ . the magnetization will be

A

$\frac{32}{3} Am^{-1}$

B

$\frac{2}{3} Am^{-1}$

C

$6 Am^{-1}$

D

$2.4 Am^{-1}$

Solution:

$I \propto \frac{B}{T}$
$\frac{I_{2}}{I_{1}}=\frac{B_{2}}{B_{1}} \times \frac{T_{1}}{T_{2}}$
$I_{2}=\frac{2}{3} A / m$

Q29. The ratio of magnetic field at the centre of a current carrying circular coil to its magnetic moment is ‘x’ if the current and the radius both are doubled. The new ratio will become

A

2x

B

4x

C

$\frac{x}{4}$

D

$\frac{x}{8}$

Solution:

Magnetic field at the center of the current carrying loop is given by
$B =$ $\left(\mu_{0} / 4 \pi\right) \times(2 \pi i / a )=\left(\mu_{0} i \right) / 2 a .$
Magnetic moment at the center of the current carrying loop is given by
$M=\pi a ^{2}$
thus $B / M =\mu_{0} /\left(2 \pi a ^{3}\right)= x ($ given $)$
when both current and radius are doubled ratio become $x / 8$ times.

Q30. In a permanent magnet at room temperature

A

Magnetic moment of each molecule is zero

B

The individual molecules have non-zero magnetic moment which are all perfectly aligned

C

Domains are partially aligned

D

Domains are all perfectly aligned

Solution:

A permanent magnet at room temperature retains a ferromagnetic property that possesses a dipole moment. And domains are partially aligned due to thermal agitation.

Q31. A rod of length $2 \,m$ slides with a speed of $5 \,ms^{-1}$ on a rectangular conducting frame as shown in figure. There exists a uniform magnetic filed of $0.04 \,T$ perpendicular to the plane of the figure. If the resistance of the rod is $3 \Omega$ . The current through the rod is

A

75 mA

B

133 mA

C

0.75 A

D

1.33 A

Solution:

We know that the current,
$i = BIv / R =(0.04 \times 2 \times 5) / 3=133\, mA$

Q32. The current in a coil of inductance $0.2 \,H$ changes from $5A$ to $2A$ in $0.5\,sec$ . The magnitude of the average induced emf in the coil is

A

0.6 V

B

1.2 V

C

30 V

D

0.3 V

Solution:

Average induced emf, $\epsilon= L dl / dt$
$=0.2((5-2) / 0.5)=2 / 5 \times 3=1.2 V$

Q33. In the given circuit the peak voltage across $C, L$ and $R$ are $30 \,V$ , $110 \,V$ and $60\, V$ respectively. The rms value of the applied voltage is

A

100 V

B

200 V

C

70.7 V

D

141 V

Solution:

$V_{0}=\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}}$
$=\sqrt{60^{2}+\left(110^{2}-30^{2}\right)}=100$
$V_{r m s}=\frac{V_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}}$
$=\frac{100}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=100 \times \frac{\sqrt{2}}{2}$
$=50 \times 1.414=70.7 \,V$

Q34. The power factor of R-L circuit is $\frac{1}{\sqrt{3}}$ . If the inductive reactance is $2\Omega$ . The value of resistance is

A

$2\Omega$

B

$\sqrt{2}\Omega$

C

$0.5\Omega$

D

$\frac{1}{\sqrt{2}}\Omega$

Solution:

$\cos \phi=\frac{1}{\sqrt{3}}, \tan \phi=\frac{\sqrt{2}}{1} \tan \phi$
$=\frac{X_{L}}{R} \sqrt{2}=\frac{2}{R}$
$\Rightarrow R=\sqrt{2} \Omega$

Q35. In the given circuit, the resonant frequency is

A

15.92 Hz

B

159.2 Hz

C

1592 Hz

D

15910 Hz

Solution:

Resonant frequency of an circuit is given by
$f=\frac{1}{2 \pi \sqrt{L C}}$
$=\frac{1}{2 \pi \sqrt{0.5 \times 10^{-3} H \times 20 \times 10^{-6} F}}$
$=1592\, Hz$

Q36. A light beam of intensity $20\, W/cm^2$ is incident normally on a perfectly reflecting surface of sides $25 \,cm \times 15 \,cm$ . The momentum imparted to the surface by the light per second is

A

$2\times10^{-5} kg \,ms^{-1}$

B

$1\times10^{-5} kg\, ms^{-1}$

C

$5\times10^{-5} kg \,ms^{-1}$

D

$1.2\times10^{-5} kg\, ms^{-1}$

Solution:

Intensity $I = E / A$
where $E$ is energy of radiation and $A$ is incident area
$\Rightarrow E = IA$
Momentum of radiation is given by
$P =2 E / c =2 IA / c$
Where $c$ is speed of light.
$P=\frac{2 \times 20 \times 10^{4} \frac{ W }{ m ^{2}} \times 375 \times 10^{-4} m ^{2}}{3 \times 10^{8} m / s }$
$=5 \times 10^{-5} kg\,ms ^{-1}$

Q37. An object approaches a convergent lens from the left of the lens with a uniform speed $5 \,m/s$ and stops at the focus, the image

A

Moves away from the lens with an uniform speed 5 m/s

B

Moves away from the lens with an uniform acceleration

C

Moves away from the lens with a non- uniform acceleration

D

Moves towards the lens with a non-uniform acceleration

Solution:

Velocity of image in convex lens $V _{ i }=( f /( f + u ))^{2} V_o$
and image will move from focus to infinity (i.e. away from the lens).
From the formula, velocity of image
$\left( dV _{ i }\right) / dt =\left( d V _{ i } / du \right) \times( du / dt )$
$=2( f / f + u ) f \ln ( f + u ) \times 5$ (because $\left.du / dt =5 m / s \right)$
$\Rightarrow \left( dV _{ i }\right) / dt =10 f ^{2} \ln ( f + u ) /( f + u )$
We can see that acceleration of image $\propto \ln (f+u)$ ,
i.e varies with distance of object
So the image moves away from the lens with a non-uniform acceleration

Q38. The refracting angle of prism is A and refractive index of material of prism is $\cot \frac{A}{2}$ The angle of minimum deviation is

A

$180^\circ -3 A$

B

$180^\circ +2 A$

C

$90^\circ - A$

D

$180^\circ -2 A$

Solution:

$n=\frac{\sin \left(\frac{A+d_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\Rightarrow \cot \left(\frac{A}{2}\right)=\frac{\sin \left(\frac{A+d_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\Rightarrow \frac{\cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\frac{\sin \left(\frac{A+d_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\Rightarrow \sin \left(90-\frac{A}{2}\right)=\sin \left(\frac{A+d_{m}}{2}\right)$
$\Rightarrow 90-\frac{A}{2}=\frac{A+d_{m}}{2}$
$\Rightarrow 180-2 A=d_{m}$

Q39. The following figure shows a beam of light converging at point $P$ . When a concave lens of focal length $16 \,cm$ is introduced in the path of the beam at a place shown by dotted line such that $OP$ becomes the axis of the lens, the beam converges at a distance $x$ from the lens. The value of $x$ will be equal to

A

12 cm

B

24 cm

C

36 cm

D

48 cm

Solution:

So, here when we put the concave lens,
let the beam will converge at a distance $x=v$
Using lens formulae, we have, $1 / f =1 / v -1 / u$
Where $u =12\, cm$ and $f =-16\, cm$ is given
$\therefore 1 / v =(1 / f )+(1 / u )$
$=(-1 / 16)+(1 / 12)=1 / 48\, cm$
$\Rightarrow v =48 \,cm$
Hence, $x=48 \,cm$

Q40. Three polaroid sheets $P_1, P_2$ and $P_3$ are kept parallel to each other such that the angle between pass axes of $P_1$ and $P_2$ is $45^\circ$ and that between $P_2$ and $P_3$ is $45^\circ$ . If unpolarised beam of light of intensity $128 Wm^{-2}$ is incident on $P_1$ . What is the intensity of light coming out of $P_3$ ?

A

$128 \,Wm^{-2}$

B

0

C

$16 \,Wm^{-2}$

D

$64\, Wm^{-2}$

Solution:

We have unpolarised beam's intensity
$I _{0}=128 w / m ^{2}$
Using Malu's law we have
$I=I_{0} \cos ^{2} \theta$
When beam passed from the first polaroid,
$I=I_{0} / 2$
Again, as the angle between $p _{1}$ and $p _{2}$ is $45^{\circ}$ ,
beam intensity when it will pass $p _{2}$ would be
$l _{1}= I _{0} / 2 \cos ^{2} 45^{0}= I _{0} / 4$
And, also the angle between $p _{2}$ and $p _{3}$ is $45^{\circ}$ ,
beam intensity when it will come out of $p _{3}$ will be
$l _{2}=I_{0} / 4 \cos ^{2} 45^{0}=I_{0} / 8$
$=128 / 8=16 w / m ^{2}$

Q41. Two poles are separated by a distance of $3.14\, m$ . The resolving power of human eye is $1$ minute of an arc. The maximum distance from which he can identify the two poles distinctly is

A

10.8 km

B

5.4 km

C

188 m

D

376 m

Solution:

Given the lateral separation of the poles $d =3.14\, m$
The resolving power of the eye is
$\theta=1 \min =\pi /(60 \times 180)$ rad
The maximum distance from which poles are distinctly visible is
$D = d / \theta$
$\Rightarrow D =(3.14 \times 60 \times 180) / \pi$
$=10800 \,m =10.8 \,km$

Q42. In young’s Double Slit Experiment, the distance between the slits and the screen is 1.2 m and the distance between the two slits is 2.4 mm. If a thin transparent mica sheet of thickness $1 \mu m$ and R.I. 1.5 is introduced between one of the interfering beams, the shift in the position of central bright fringe is

A

2 mm

B

0.5 mm

C

0.125 mm

D

0.25 mm

Solution:

Path difference due to insertion of mica sheet $\Delta x =(\mu-1) t$
Let the shift in the fringe pattern be $'y'$
Also, path difference $\Delta x=y \times d / D$ ,
so comparing both $(\mu-1) t=y \times(d / D)$
$y =(\mu-1) t \times( D / d )$ ,
where $\mu=1.5, D =2.4$ and $d =1.2$
putting the values, we get
$y =0.25 \,mm$ .

Q43. The de-Broglie wavelength associated with electron of hydrogen atom in this ground state is

A

0.3 $\mathring{A}$

B

3.3 $\mathring{A}$

C

6.26 $\mathring{A}$

D

10 $\mathring{A}$

Solution:

De-Broglie wavelength is given by
$\lambda=12.27 / \sqrt{ E }_{0}$ ,
where $E _{0}$ is the ground state energy of the hydrogen atom
whose value is $13.6\, V$
$\therefore \lambda=12.27 /(\sqrt{1} 3.6)=3.33 \,\mathring{A}$

Q44. The following graph represents the variation of photo current with anode potential for a metal surface. Here $I_1$ , $I_2$ and $I_3$ represents intensities and $\gamma_1$ , $\gamma_2$ , $\gamma_3$ represent frequency for curves $1, 2$ and $3$ respectively, then

A

$\gamma_1 = \gamma_2$ and $I_1 \ne I_2$

B

$\gamma_1 = \gamma_3$ and $I_1 \ne I_3$

C

$\gamma_1 = \gamma_2$ and $I_1 = I_2$

D

$\gamma_2 = \gamma_3$ and $I_1 = I_3$

Solution:

Here in the graph we can see that, the Stopping potential is same for $1$ and $2$ . So, frequencies will be same i.e. $\gamma _{1}= \gamma _{2}$ and, currents are different. So, intensity are different i.e. $I_{1} \neq I_{2}$ .

Q45. The period of revolution of an electron revolving in $n^{th}$ orbit of $H$ -atom is proportional to

A

$n^2$

B

$\frac{1}{n}$

C

$n^3$

D

Independent of n

Solution:

As we know that time period of revolution of an electron is
$T =2 \pi r / v$ ,
And $r \propto n^{2} / Z^{2}$ and $v \propto Z / n$
$\therefore T \propto n ^{3} / Z ^{2}$ and for $H$ -atom
$Z =1, T \propto n ^{3}$

Q46. Angular momentum of an electron in hydrogen atom is $\frac{3h}{2\pi}$ (h is the Planck’s constant). The K.E. of the electron is

A

4.35 eV

B

1.51 eV

C

3.4 eV

D

6.8 eV

Solution:

Angular momentum,
$L =\frac{ nh }{2 \pi}=\frac{3 h }{2 \pi}$
$given$
also $n =3$
$\therefore E =\frac{-1.36}{ n ^{2}} eV$ $E =\frac{-13.6}{(3)^{2}} eV$
$E =\frac{-13.6}{9} eV$
$E = -1.57 \,eV$

Q47. A beam of fast moving alpha particles were directed towards a thin film of gold. The parts A, B and C of the transmitted and reflected beams corresponding to the incident parts A, B and C of the beam are shown in the adjoining diagram. The number of alpha particles in

A

C' will be minimum and in B' maximum

B

B' will be minimum and in C' maximum

C

A' will be maximum and in B' minimum

D

A' will be minimum and in B' maximum

Solution:

$A'$ will be maximum and $B'$ will be minimum, because atom is hollow and whole mass of the atom is concentrated in a small centre called nucleus.

Q48. Two protons are kept at a separation of $10 \,nm$ . Let $F_n$ and $F_e$ the nuclear force and the electromagnetic force between them

A

$F_e = F_n$

B

$F_e >> F_n$

C

$F_e << F_n$

D

$F_e$ and $F_n$ differ only slightly

Solution:

Nuclear forces are short range force existing in the range of a $10^{-15} m$ .
If the separation between the particles is greater than $10^{-15} m$ ,
nuclear force is negligible.
Therefore, at a separation of $10 \,nm$ ,
the electromagnetic force is greater than the nuclear force.
Hence, $F _{ e } > > F _{ n }$

Q49. During a $\beta^-$ decay

A

An atomic electron is ejected

B

An electron which is already present within the nucleus is ejected

C

A neutron in the nucleus decays emitting an electron

D

A proton in the nucleus decays emitting an electron

Solution:

During $β^-$ decay, a neutron in the nucleus decays emitting an electron.

Q50. A radio-active elements has half-life of $15$ years. What is the fraction that will decay in $30$ years?

A

0.25

B

0.5

C

0.75

D

0.85

Solution:

Fraction un decayed is given as,
$N / N _{0}=(1 / 2)^{( t / T )}$
Here, $t =30$ years and $T =15$ years
So,
$N / N _{0}=(1 / 2)^{((30 \text { years }) /(15 \text { years }))}$
or, $N / N _{0}=(1 / 2)^{2}$
or, $N / N _{0}=1 / 4$
Therefore,
Fraction un decayed $=1- N / N _{0}$
or, Fraction un decayed $=1-(1 / 4)$
or, Fraction un decayed $=3 / 4=0.75$

Q51. A $220\, V$ A.C supply is connected between points $A$ and $B$ as shown in figure what will be the potential difference $V$ across the capacitor?

A

220 V

B

110 V

C

0

D

$220\sqrt{2} V$

Solution:

During the positive half cycles of input ac, a $p-n$ junction diode is forward biased and hence it conducts.
While during the negative half cycles, the diode will be reverse biased and hence does not conduct.
The capacitor is charged to maximum potential difference.
Therefore, the potential difference across capacitor $C$ is equal to the peak value of the applied ac voltage, i.e..
$V = V _{\max }$
or, $V =\sqrt{2} V _{ rms }$
or, $V =\sqrt{2} \times 220 \,V =200 \sqrt{2}\, V$

Q52. In the following circuit what are $P$ and $Q$ :

A

P = 1, Q = 0

B

P = 0, Q = 1

C

P = 0, Q = 0

D

P = 1, Q = 1

Solution:

A $NAND$ gate gives an output $1$ if at least one of the inputs is zero.
Hence $Q$ is $1$ .
Therefore, the inputs for the upper $NAND$ gate are $1,1$ .
Hence $P =0$ .

Q53. A positive hole in a semiconductor is

A

An anti-particle of electron

B

A vacancy created when an electron leaves a covalent bond

C

Absence of free electrons

D

An artificially created particle

Solution:

A positive hole in a semiconductor is a vacancy which is created at the site of a covalent bond when an electron leaves a covalent bond.

Q54. Two long straight parallel wires are a distance $2 d$ apart. They carry steady equal currents flowing out of the plane of the paper. The variation of magnetic field $B$ along the line $xx'$ is given by

A

B

C

D

Solution:

The magnetic field due to a long straight current carrying wire is given by,
$B =\left(\mu_{0} I \right) / 2 \pi r$
or, $B \propto 1 / r$
At the point exactly mid-way between the conductors, the net magnetic field is zero. Using right hand thumb rule, we find that the magnetic field due to left wire will be in $\hat{j}$ direction while due to the right wire is in $(-\hat{j})$ direction.
Magnetic field at a distance $x$ from the left wire, lying between the wires.
$B=\frac{\mu_{0} I}{2 \pi x} \hat{j}+\frac{\mu_{0} I}{2 \pi(2 d-x)}(-\hat{j})$
or, $B=\frac{\mu_{0} I}{2 \pi}\left(\frac{1}{x}-\frac{1}{2 d-x}\right)$
At $x=d, B=0$
For $x < d, B$ is along $\hat{j}$
For $x > d, B$ is along $-\hat{j}$
On the left side of the left conductor, magnetic fields due to the currents will add up and the net magnetic field will be along $(-\hat{j})$ direction.
To the right side of second conductor, the total magnetic field will be along $\hat{j}$ direction.

Q55. A cylindrical wire has a mass $(0.3 \pm 0.003) g$ , radius $(0.5 \pm 0.005) mm$ and length $(6 \pm 0.06) cm$ . The maximum percentage error in the measurement of its density is

A

1

B

2

C

3

D

4

Solution:

$\rho=\frac{m}{V}=\frac{m}{\pi r^{2} l}$
or, $\frac{\Delta \rho}{\rho}=\frac{\Delta m}{m}+2 \frac{\Delta r }{r}+\frac{\Delta l}{l}$
or, $\frac{\Delta \rho}{\rho} \times 100=\left(\frac{0.003}{0.3}+2 \frac{0.005}{0.5}+\frac{0.06}{6}\right) \times 100$
or, $\frac{\Delta \rho}{\rho} \times 100=(0.01+0.02+0.01) \times 100$
or, $\frac{\Delta \rho}{\rho} \times 100=0.04 \times 100$
$(\Delta \rho / \rho) \times 100=4 \%$
Therefore, the percentage error in the measurement of density is $4 .$

Q56. At a metro station, a girl walks up a stationary escalator in $20 \,sec$ . If she remains stationary on the escalator, then the escalator take her up in $30 \,sec$ . The time taken by her to walk up on the moving escalator will be

A

25 sec

B

60 sec

C

12 sec

D

10 sec

Solution:

Let, the length of the escalator be $L$ .
Velocity of the girl, $v_{g}=L / t_{1}=L / 20$
Velocity of the escalator, $v _{ e }= L / t _{2}= L / 30$
On moving escalator.
$v_{\text {net }}=v_{g}+v_{e}$
or, $v _{\text {net }}=( L / 20)+( L / 30)$
or, $v _{\text {net }}= L / 12$
Since, $v_{\text {net }}=L / t$
Therefore,
$L / t = L / 12$
$t =12 s$

Q57. Rain is falling vertically with a speed of $12 \,ms^{-1}$ . A woman rides a bicycles with a speed of $12 \,ms^{-1}$ in east to west direction. What is the direction in which she should hold her umbrella?

A

$30^\circ$ towards East

B

$45^\circ$ towards East

C

$30^\circ$ towards West

D

$45^\circ$ towards West

Solution:

To protect herself from the rain, the woman must hold her umbrella in the direction of the relative velocity of the rain with respect to the woman.
$\tan\, \theta= v _{ w } / v _{ f }=12\, m / s$
or, $\tan \,\theta=1$
or, $\theta=45^{\circ}$
Therefore, the direction in which she should hold her umbrella is $45^{\circ}$ toward west.

Q58. One end of a string of length $'l'$ is connected to a particle of mass ‘m’ and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed ‘v’, the net force on the particle (directed towards the centre) is : (T is the tension in the string)

A

T

B

$T-\frac{mv^2}{1}$

C

$T+\frac{mv^2}{1}$

D

0

Solution:

When a particle connected to a string revolves in a circular path around a centre, the centripetal force is provided by the tension produced in the string. Hence, in the given case, the net force on the particle is the tension $T$ . $\left(T=\left( mv ^{2}\right) / I \right)$

Q59. A body is initially at rest. It undergoes one- dimensional motion with constant acceleration. The power delivered to it at time ‘t’ is proportional to

A

$t^{1/2}$

B

t

C

$t^{3/2}$

D

$t^{2}$

Solution:

As we know,
$P = Fv$
where, $P$ is the power, $F$ is the force and $v$ is the velocity.
or, $P =$ mav
or, $P = ma ^{2} t$
or, $P \propto t$

Q60. A thin uniform rectangular plate of mass $2\, kg$ is placed in $X-Y$ plane as shown in figure. The moment of inertial about x-axis is $I_x = 0.2 \,kg \,m^2$ and the moment of inertia about y-axis is $I_y = 0.3\, kg\, m^2$ . The radius of gyration of the plate about the axis passing through O and perpendicular to the plane of the plate is

A

38.7 cm

B

50 cm

C

38.8 cm

D

31.6 cm

Solution:

By perpendicular axis theorem we know that,
$I _{ z }= I _{ x }+ l _{ y }=0.3+0.2=0.5 \,kgm ^{2}$
as we know that MOI
$I = mK ^{2} \Rightarrow 0.5=2 \times K ^{2}$
$K =1 / 2=0.5 m =50 \,cm$

KCET 2020 Physics Questions with Answers Key Solutions

Q1. The value of acceleration due to gravity at a height of 10km10\,km from the surface of earth is xx. At what depth inside the earth is the value of the acceleration due to gravity has the same value xx?

5 km

20 km

10 km

15 km

gh=g(1−2hR)g_{h}=g\left(1-\frac{2 h}{R}\right) gd=g(1−dR)g_{d}=g\left(1-\frac{d}{R}\right) ⇒gh=gd\Rightarrow g_{h}=g_{d} g(1−2hR)=g(1−dR)g\left(1-\frac{2 h}{R}\right)=g\left(1-\frac{d}{R}\right) ⇒d=2h=2×10=20km\Rightarrow d=2 h=2 \times 10=20\, km

Q2. Young’s modulus of a perfect rigid body is

Zero

Unity

Infinity

Between zero and unity

Young’s modulus = stress/strain, and for a perfect rigid body we know that strain is equal to 00 and hence Young’s modulus of a perfect rigid body becomes Infinity.

Q3. A wheel starting from rest gains an angular velocity of 1010 rad/s after uniformly accelerated for 55 sec. The total angle through which it has turned is

25 rad

100 rad

25 π\pi rad

50 π\pi rad about a vertical axis

ω1=0\omega_{1}=0 ω2=10radsect=5sec\omega_{2}=10 \frac{ rad }{ sec } t=5 \,sec θ=ω1+ω22×t\theta=\frac{\omega_{1}+\omega_{2}}{2} \times t =102×5=25rad=\frac{10}{2} \times 5=25 \,rad

Q4. Iceberg floats in water with part of it submerged. What is the fraction of the volume of iceberg submerged if the density of ice is ρi=0.917gcm−3\rho_i = 0.917 \,g\,cm^{-3} ?

0.917

1

0.458

0

Q5. A sphere, a cube and a thin circular plate all of same material and same mass initially heated to same high temperature are allowed to cool down under similar conditions. Then the

plate will cool the fastest and cube the slowest

sphere will cool the fastest and cube the slowest

plate will cool the fastest and sphere the slowest

cube will cool the fastest and plate the slowest

Surface area is more for plate and less for sphere. Hence plate will cool the fastest and sphere the slowest.

Q6. In an adiabatic expansion of an ideal gas the product of pressure and volume

Decreases

Increases

Remains constant

At first increases and then decreases

In an adiabatic expansion as temperature decreases from an ideal gas equation PV=nRTPV=nRT. Product of pressure and volume also decreases.

Q7. A certain amount of heat energy is supplied to a monoatomic ideal gas which expands at constant pressure. What fraction of the heat energy is converted into work ?

1

23\frac {2}{3}

25\frac {2}{5}

57\frac {5}{7}

Q8. A tray of mass 12kg12 \,kg is supported by two identical springs as shown in figure. When the tray is pressed down slightly and then released, it executes SHM with a time period of 1.5s1.5\,s. The spring constant of each spring is

50Nm−150 \,Nm^{-1}

00

105Nm−1105 \,Nm^{- 1}

∞\infty

We know that T=2π√mkeffT=2 \pi \sqrt{\frac{m}{k_{\text{eff}}}} 32=2π√122k\frac{3}{2}=2 \pi \sqrt{\frac{12}{2 k}} ⇒94=4π2×122kk≈105Nm−1 \Rightarrow \frac{9}{4}=4 \pi^{2} \times \frac{12}{2 k} k \approx 105 \,Nm ^{-1}

Q9. A train whistling at constant frequency ‘nn’ is moving towards a station at a constant speed VV. The train goes past a stationary observer on the station. The frequency ‘nn’ of the sound as heard by the observer is plotted as a function of time ‘tt’. Identify the correct curve

Q10. A point charge ‘qq’ is placed at the corner of a cube of side ‘aa’ as shown in the figure. What is the electric flux through the face ABCDABCD ?

00

q24ϵo\frac {q}{24\epsilon_o}

q6ϵo\frac {q}{6\epsilon_o}

q72ϵo\frac {q}{72\epsilon_o}

Q11. The electric field lines on the left have twice the separation on those on the right as shown in figure. If the magnitude of the field at AA is 40Vm−140\,Vm^{-1}, what is the force on 20μC20\,\mu \,C charge kept at BB?

4×10−4Vm−14 \times 10^{-4} Vm^{-1}

8×10−4Vm−18 \times 10^{-4} Vm^{-1}

16×10−4Vm−116 \times 10^{-4} Vm^{-1}

1×10−4Vm−11 \times 10^{-4} Vm^{-1}

We know that F=qE=20×20×10−6F = qE =20 \times 20 \times 10^{-6} =4×10−4V/m−1=4 \times 10^{-4} V / m^{-1}

Q12. An infinitely long thin straight wire has uniform charge density of 14×10−2cm−1.\frac {1}{4} \times 10^{-2} cm^{-1}. What is the magnitude of electric field at a distance 20cm20\, cm from the axis of the wire ?

1.12×108NC−11.12 \times 10^8 \,NC^{-1}

4.5×108NC−14.5 \times 10^8 \,NC^{-1}

2.25×108NC−12.25 \times 10^8 \,NC^{-1}

9×108NC−19 \times 10^8 \,NC^{-1}

We know that E=λ2πε0rE=\frac{\lambda}{2 \pi \varepsilon_{0} r} =14×10−210−2×18×109×5=\frac{1}{4} \times \frac{10^{-2}}{10^{-2}} \times 18 \times 10^{9} \times 5 =2.25×108N/C=2.25 \times 10^{8} N / C

Q13. A dipole moment ‘PP’ and moment of inertia II is placed in a uniform electric field →E\vec{E}. If it is displaced slightly from its stable equilibrium position, the period of oscillation of dipole is

√PEI\sqrt{\frac{PE}{I}}

2π√IPE2\pi\sqrt{\frac{I}{PE}}

12π√PEI \frac {1}{2\pi} \sqrt{\frac{PE}{I}}

π√IPE \pi \sqrt{\frac{I}{PE}}

Q14. The difference between equivalent capacitances of two identical capacitors connected in parallel to that in series is 6μF6 \,\mu F . The value of capacitance of each capacitor is

2μF2 \mu F

3μF3 \mu F

4μF4 \mu F

6μF6 \mu F

CP−C=6μFC _{ P - C }=6\, \mu F 2C−(C/2)=62 C -( C / 2)=6 C=4μFC =4\, \mu F

VA=VB=VCV_A = V_B = V_C

VA=VB>VCV_A = V_B > V_C

VA=VB<VCV_A = V_B < V_C

VA>VB=VCV_A > V_B = V_C

Q1. The value of acceleration due to gravity at a height of $10\,km$ from the surface of earth is $x$ . At what depth inside the earth is the value of the acceleration due to gravity has the same value $x$ ?

$g_{h}=g\left(1-\frac{2 h}{R}\right)$
$g_{d}=g\left(1-\frac{d}{R}\right)$
$\Rightarrow g_{h}=g_{d}$
$g\left(1-\frac{2 h}{R}\right)=g\left(1-\frac{d}{R}\right)$
$\Rightarrow d=2 h=2 \times 10=20\, km$

Young’s modulus = stress/strain, and for a perfect rigid body we know that strain is equal to $0$ and hence Young’s modulus of a perfect rigid body becomes Infinity.

Q3. A wheel starting from rest gains an angular velocity of $10$ rad/s after uniformly accelerated for $5$ sec. The total angle through which it has turned is

25 $\pi$ rad

50 $\pi$ rad about a vertical axis

$\omega_{1}=0$
$\omega_{2}=10 \frac{ rad }{ sec } t=5 \,sec$
$\theta=\frac{\omega_{1}+\omega_{2}}{2} \times t$
$=\frac{10}{2} \times 5=25 \,rad$

Q4. Iceberg floats in water with part of it submerged. What is the fraction of the volume of iceberg submerged if the density of ice is
$\rho_i = 0.917 \,g\,cm^{-3}$ ?

In an adiabatic expansion as temperature decreases from an ideal gas equation $PV=nRT$ .
Product of pressure and volume also decreases.

$\frac {2}{3}$

$\frac {2}{5}$

$\frac {5}{7}$

Q8. A tray of mass $12 \,kg$ is supported by two identical springs as shown in figure. When the tray is pressed down slightly and then released, it executes SHM with a time period of $1.5\,s$ . The spring constant of each spring is

$50 \,Nm^{-1}$

$0$

$105 \,Nm^{- 1}$

$\infty$

We know that
$T=2 \pi \sqrt{\frac{m}{k_{\text{eff}}}}$
$\frac{3}{2}=2 \pi \sqrt{\frac{12}{2 k}}$
$\Rightarrow \frac{9}{4}=4 \pi^{2} \times \frac{12}{2 k} k \approx 105 \,Nm ^{-1}$

Q10. A point charge ‘ $q$ ’ is placed at the corner of a cube of side ‘ $a$ ’ as shown in the figure. What is the electric flux through the face $ABCD$ ?

$0$

$\frac {q}{24\epsilon_o}$

$\frac {q}{6\epsilon_o}$

$\frac {q}{72\epsilon_o}$

Q11. The electric field lines on the left have twice the separation on those on the right as shown in figure. If the magnitude of the field at $A$ is $40\,Vm^{-1}$ , what is the force on $20\,\mu \,C$ charge kept at $B$ ?

$4 \times 10^{-4} Vm^{-1}$

$8 \times 10^{-4} Vm^{-1}$

$16 \times 10^{-4} Vm^{-1}$

$1 \times 10^{-4} Vm^{-1}$

We know that
$F = qE =20 \times 20 \times 10^{-6}$
$=4 \times 10^{-4} V / m^{-1}$

Q12. An infinitely long thin straight wire has uniform charge density of $\frac {1}{4} \times 10^{-2} cm^{-1}.$ What is the magnitude of electric field at a distance $20\, cm$ from the axis of the wire ?

$1.12 \times 10^8 \,NC^{-1}$

$4.5 \times 10^8 \,NC^{-1}$

$2.25 \times 10^8 \,NC^{-1}$

$9 \times 10^8 \,NC^{-1}$

We know that
$E=\frac{\lambda}{2 \pi \varepsilon_{0} r}$
$=\frac{1}{4} \times \frac{10^{-2}}{10^{-2}} \times 18 \times 10^{9} \times 5$
$=2.25 \times 10^{8} N / C$

Q13. A dipole moment ‘ $P$ ’ and moment of inertia $I$ is placed in a uniform electric field $\vec{E}$ . If it is displaced slightly from its stable equilibrium position, the period of oscillation of dipole is

$\sqrt{\frac{PE}{I}}$

$2\pi\sqrt{\frac{I}{PE}}$

$\frac {1}{2\pi} \sqrt{\frac{PE}{I}}$

$\pi \sqrt{\frac{I}{PE}}$

Q14. The difference between equivalent capacitances of two identical capacitors connected in parallel to that in series is $6 \,\mu F$ . The value of capacitance of each capacitor is

$2 \mu F$

$3 \mu F$

$4 \mu F$

$6 \mu F$

$C _{ P - C }=6\, \mu F$
$2 C -( C / 2)=6$
$C =4\, \mu F$

$V_A = V_B = V_C$

$V_A = V_B > V_C$

$V_A = V_B < V_C$

$V_A > V_B = V_C$

Electric lines of force in an electric field always flow from a higher potential to a lower potential.
Hence, $V _{ A }= V _{ B } > V _{ C }$

Q17. A hot filament liberates an electron with zero initial velocity. The anode potential is $1200\,V$ . The speed of the electron when it strikes the anode is

$1.5 \times 10^5 \,ms^{-1}$

$2.5 \times 10^6 \,ms^{-1}$

$2.1 \times 10^7 \,ms^{-1}$

$2.5 \times 10^8 \,ms^{-1}$

$\frac{1}{2} m v^{2}=V q$
$v=\sqrt{\frac{2 V q}{m}}$
$=2.1 \times 10^{7} ms ^{-1}$

Q18. A metal rod of length $10 \,cm$ and a rectangular cross - section of $1 \,cm \times \frac{1}{2}\, cm$ is connected to a battery across opposite faces. The resistance will be

maximum when the battery is connected across $1 \,cm \times \frac {1}{2}$ cm faces

maximum when the battery is connected across $10 \,cm\, \times \frac {1}{2} cm$ faces

maximum when the battery is connected across $10 \,cm \times$ 1 cm faces

Resistance is inversely proportional to area.
Hence resistance will be maximum when the battery is connected across $1 \,cm \times$ $(1 / 2) cm$ faces.

Q19. A car has a fresh storage battery of e.m.f $12\,V$ and internal resistance $2 \times 10^{-2} \Omega$ . If the starter motor draws a current of $80\,A$ . Then the terminal voltage when the starter is on is

$V=E-ir=10.4 \,V$

Q20. A potentiometer has a uniform wire of length $5\,m$ . A battery of emf $10\,V$ and negligible internal resistance is connected between its ends. A secondary cell connected to the circuit gives balancing length at $200 \,cm$ . The emf of the secondary cell is

Emf of cell in the secondary circuit =potential gradient $x$ balancing length
$\Rightarrow (10 / 5) \times 2=4\, V$

Q21. The colour code for a carbon resistor of resistance $0.28\,k\Omega \pm 10$ % is

Q22. Each resistance in the given cubical network has resistance of $1\,\Omega$ and equivalent resistance between $A$ and $B$ is

$\frac {5} {6} \Omega$

$\frac {6} {5} \Omega$

$\frac {5} {12} \Omega$

$\frac {12} {5} \Omega$

$5 / 6\, \Omega$

Q23. $I-V$ characteristic of a copper wire of length $L$ and area of cross-section $A$ is shown in figure. The slope of the curve becomes

Slope $=1 / R = A / \rho l$
Hence the slope becomes less if the length of the wire is increased.

Q24. In the given figure, the magnetic field at ‘ $O$ ’.

$\frac {3}{4} \frac {\mu_oI}{r}+\frac {\mu_oI}{4\pi r}$

$\frac {3}{10} \frac {\mu_oI}{r} - \frac {\mu_oI}{4\pi r}$

$\frac {3}{8} \frac {\mu_oI}{r} + \frac {\mu_oI}{4\pi r}$

$\frac {3}{8} \frac {\mu_oI}{r} - \frac {\mu_oI}{4\pi r}$