KCET 2020 Physics Questions with Answers Key Solutions
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) $
$\Rightarrow d=2 h=2 \times 10=20\, km$
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.
Solution:
$\omega_{1}=0$
$\omega_{2}=10 \frac{ rad }{ sec } t=5 \,sec$
$\theta=\frac{\omega_{1}+\omega_{2}}{2} \times t$
$=\frac{10}{2} \times 5=25 \,rad$
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 $
$\Rightarrow V' / V=\rho_{i} / \rho_{w}$
$=0.917 / 1=0.917$
Solution:
Surface area is more for plate and less for sphere. Hence plate will cool the fastest and sphere the slowest.
Solution:
In an adiabatic expansion as temperature decreases from an ideal gas equation $PV=nRT$.
Product of pressure and volume also decreases.
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} $
$\frac{c_{p}-c_{v}}{c_{p}}=1-\frac{1}{\gamma}$
$=1-\frac{1}{\frac{5}{3}}=\frac{2}{5}$
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}$
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}
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 $\frac{ q }{24 \epsilon_{0}}$;
Solution:
We know that
$F = qE =20 \times 20 \times 10^{-6}$
$=4 \times 10^{-4} V / m^{-1}$
Solution:
We know that
$E=\frac{\lambda}{2 \pi \varepsilon_{0} r}$
$=\frac{1}{4} \times \frac{10^{-2}}{10^{-2}} \times 18 \times 10^{9} \times 5$
$=2.25 \times 10^{8} N / C$
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)
Solution:
$C _{ P - C }=6\, \mu F$
$2 C -( C / 2)=6$
$C =4\, \mu F$
Solution:
Electric lines of force in an electric field always flow from a higher potential to a lower potential.
Hence, $V _{ A }= V _{ B } > V _{ C }$
Solution:
When the radius of a soap bubble increases, the soap bubble gets charged.
Solution:
$\frac{1}{2} m v^{2}=V q$
$v=\sqrt{\frac{2 V q}{m}}$
$=2.1 \times 10^{7} ms ^{-1}$
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.
Solution:
$V=E-ir=10.4 \,V$
Solution:
Emf of cell in the secondary circuit =potential gradient $x$ balancing length
$\Rightarrow (10 / 5) \times 2=4\, V$
Solution:
Red, Grey, Brown, Silver
Solution:
$5 / 6\, \Omega$
Solution:
Slope $=1 / R = A / \rho l$
Hence the slope becomes less if the length of the wire is increased.
Solution:
$B_{\text {net }}=B_{1}+B_{2}+B_{3}$
$=\frac{3}{8} \frac{\mu_{0} I}{r}+\frac{\mu_{0} I}{4 \pi r}+0$
Solution:
$\frac{\mu_{0} i}{4 \pi} \frac{\overrightarrow{d l} \times \vec{r}}{r^{3}}$
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.
Solution:
K.E. $=q^{2} / m$
Hence, a deuteron gains the least kinetic energy.
Solution:
$I \propto \frac{B}{T}$
$\frac{I_{2}}{I_{1}}=\frac{B_{2}}{B_{1}} \times \frac{T_{1}}{T_{2}}$
$I_{2}=\frac{2}{3} A / m$
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.
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.
Solution:
We know that the current,
$i = BIv / R =(0.04 \times 2 \times 5) / 3=133\, mA$
Solution:
Average induced emf, $\epsilon= L dl / dt$
$=0.2((5-2) / 0.5)=2 / 5 \times 3=1.2 V$
Solution:
$V_{0}=\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}}$
$=\sqrt{60^{2}+\left(110^{2}-30^{2}\right)}=100$
$V_{r m s}=\frac{V_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}}$
$=\frac{100}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=100 \times \frac{\sqrt{2}}{2}$
$=50 \times 1.414=70.7 \,V$
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$
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}} $
$=1592\, Hz$
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.
$P=\frac{2 \times 20 \times 10^{4} \frac{ W }{ m ^{2}} \times 375 \times 10^{-4} m ^{2}}{3 \times 10^{8} m / s }$
$=5 \times 10^{-5} kg\,ms ^{-1}$
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
Solution:
$n=\frac{\sin \left(\frac{A+d_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\Rightarrow \cot \left(\frac{A}{2}\right)=\frac{\sin \left(\frac{A+d_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\Rightarrow \frac{\cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\frac{\sin \left(\frac{A+d_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\Rightarrow \sin \left(90-\frac{A}{2}\right)=\sin \left(\frac{A+d_{m}}{2}\right)$
$\Rightarrow 90-\frac{A}{2}=\frac{A+d_{m}}{2}$
$\Rightarrow 180-2 A=d_{m}$
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, $x=48 \,cm$
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
$l _{2}=I_{0} / 4 \cos ^{2} 45^{0}=I_{0} / 8$
$=128 / 8=16 w / m ^{2}$
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
$D = d / \theta$
$\Rightarrow D =(3.14 \times 60 \times 180) / \pi$
$=10800 \,m =10.8 \,km$
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
$y =0.25 \,mm$.
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 $13.6\, V$
$\therefore \lambda=12.27 /(\sqrt{1} 3.6)=3.33 \,\mathring{A}$
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
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
$Z =1, T \propto n ^{3}$
Solution:
Angular momentum,
$L =\frac{ nh }{2 \pi}=\frac{3 h }{2 \pi} $
$ given$
also $ n =3$
$\therefore E =\frac{-1.36}{ n ^{2}} eV$
$E =\frac{-13.6}{(3)^{2}} eV$
$E =\frac{-13.6}{9} eV$
$E = -1.57 \,eV$
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.
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, $F _{ e } > > F _{ n }$
Solution:
During $β^-$ decay, a neutron in the nucleus decays emitting an electron.
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 $=3 / 4=0.75$
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, $V =\sqrt{2} \times 220 \,V =200 \sqrt{2}\, V$
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 $P =0$.
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.
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.
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 $4 .$
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,
$L / t = L / 12$
$t =12 s$
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.
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 $T$. $\left(T=\left( mv ^{2}\right) / I \right)$
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, $P \propto t$
Solution: