Search any question & find its solution
Question:
Answered & Verified by Expert
\(P, Q\) 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
Options:
Solution:
1007 Upvotes
Verified Answer
The correct answer is:
1
Using Lami's theorem, we have
$\begin{aligned} & \frac{P}{\sin \theta_1}=\frac{R}{\sin 150^{\circ}} \Rightarrow \frac{P}{\sin \theta_1}=\frac{R}{\sin \left(180^{\circ}-30^{\circ}\right)} \\ & \Rightarrow \frac{1.9318 \mathrm{kgwt}}{0.9659}=\frac{R}{1 / 2} \\ & \Rightarrow R=\frac{1.9318}{0.9659} \times \frac{1}{2} \mathrm{kgwt}=1 \mathrm{kgwt} .\end{aligned}$
$\begin{aligned} & \frac{P}{\sin \theta_1}=\frac{R}{\sin 150^{\circ}} \Rightarrow \frac{P}{\sin \theta_1}=\frac{R}{\sin \left(180^{\circ}-30^{\circ}\right)} \\ & \Rightarrow \frac{1.9318 \mathrm{kgwt}}{0.9659}=\frac{R}{1 / 2} \\ & \Rightarrow R=\frac{1.9318}{0.9659} \times \frac{1}{2} \mathrm{kgwt}=1 \mathrm{kgwt} .\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.