P, Q and R are three coplanar forces acting at a point and are in equilibium. Given : P=1.9318 ~kg wt [sin θ_1=0.9659 ] , the value of R (in kg wt) is

Correct answer: (C) 1. Using Lami's theorem, we have P/(sin θ_1)=fracRsin 150^° ⇒ P/(sin θ_1)=fracRsin (180^°-30^° ) ⇒ (1.9318 kgwt)/0.9659=R/(1 / 2) ⇒ R=1.9318/0.9659 × 1/2 kgwt=1 kgwt .

Q142 / 180 · AIIMS 2008 · Physics

$P, Q$ and $R$ are three coplanar forces acting at a point and are in equilibium. Given : $P=1.9318 \mathrm{~kg}$ wt $\left[\sin \theta_1=0.9659\right]$, the value of $R$ (in $\mathrm{kg}$ wt) is

A
0.9659
B
2
Correct answer
C
1
D
$1 / 2$.

Solution

Using Lami's theorem, we have $\begin{aligned} & \frac{P}{\sin \theta_1}=\frac{R}{\sin 150^{\circ}} \Rightarrow \frac{P}{\sin \theta_1}=\frac{R}{\sin \left(180^{\circ}-30^{\circ}\right)} \\ & \Rightarrow \frac{1.9318 \mathrm{kgwt}}{0.9659}=\frac{R}{1 / 2} \\ & \Rightarrow R=\frac{1.9318}{0.9659} \times \frac{1}{2} \mathrm{kgwt}=1 \mathrm{kgwt} .\end{aligned}$

Hence, the correct answer is option: 'C'.

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\(P, Q\) and \(R\) are three coplanar forces acting at a point and are in equilibium. Given : \(P=1.9318 \mathrm{~kg}\) wt \(\left[\sin \theta_1=0.9659\right]\), the value of \(R\) (in \(\mathrm{kg}\) wt) is

Solution

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