A 1 N pendulum bob is held at an angle θ from the vertical by a 2 N horizontal force F as shown in the figure. The tension in the string supporting the pendulum bob (in newton) is

Question

A 1 N pendulum bob is held at an angle θ from the vertical by a 2 N horizontal force F as shown in the figure. The tension in the string supporting the pendulum bob (in newton) is (A)  cos θ (B) (2/ cos θ) (C) √5 (D) 1

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As the bob is in equilibrium under the effect of three forces

- T, F

and

W

∴ T + P + W = 0

From, the figure

T cos θ = W ... (i)

T sin θ = F ... (ii)

From Eqs. (i) and (ii)

T^{2} = W^{2} + F^{2}

or

T = W^{2} + F^{2}

= (1)^{2} + (2)^{2}

= 5

Q. $A 1 N$ pendulum bob is held at an angle $θ$ from the vertical by a $2 N$ horizontal force $F$ as shown in the figure. The tension in the string supporting the pendulum bob (in newton) is

$cos θ$

$\frac{2}{c o s θ}$

$5$

1

As the bob is in equilibrium under the effect of three forces $- T, F$ and $W$

$∴ T + P + W = 0$
From, the figure $T cos θ = W ... (i)$
$T sin θ = F ... (ii)$
From Eqs. (i) and (ii)
$T^{2} = W^{2} + F^{2}$
or $T = W^{2} + F^{2}$
$= (1)^{2} + (2)^{2}$
$= 5$

Q. A1N pendulum bob is held at an angle θ from the vertical by a 2N horizontal force F as shown in the figure. The tension in the string supporting the pendulum bob (in newton) is

cosθ

cosθ2​

5​

1

As the bob is in equilibrium under the effect of three forces −T,F and W ∴T+P+W=0 From, the figure Tcosθ=W...(i) Tsinθ=F...(ii) From Eqs. (i) and (ii) T2=W2+F2 or T=W2+F2​ =(1)2+(2)2​ =5​

Q. $A 1 N$ pendulum bob is held at an angle $θ$ from the vertical by a $2 N$ horizontal force $F$ as shown in the figure. The tension in the string supporting the pendulum bob (in newton) is

$cos θ$

$\frac{2}{c o s θ}$

$5$

As the bob is in equilibrium under the effect of three forces $- T, F$ and $W$

$∴ T + P + W = 0$
From, the figure $T cos θ = W ... (i)$
$T sin θ = F ... (ii)$
From Eqs. (i) and (ii)
$T^{2} = W^{2} + F^{2}$
or $T = W^{2} + F^{2}$
$= (1)^{2} + (2)^{2}$
$= 5$