Q. The potential energy for a force field $\overrightarrow{ F }$ is given by $U(x, y)=\sin (x+y)$. The magnitude of force acting on the particle of mass $m$ at $\left(0, \frac{\pi}{4}\right)$ is :

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A

$1$

B

$\sqrt{2}$

C

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

D

$0$

Solution:

We use
$\overrightarrow{ F } =\nabla U =-\cos (x+y) \hat{ i }-\cos (x+y) \hat{ j }$
$\left.\overrightarrow{ F }\right|_{\left(0, \frac{\pi}{4}\right)} =-\frac{1}{\sqrt{2}} \hat{ i }-\frac{1}{\sqrt{2}} \hat{ j } $
$|\overrightarrow{ F }|_{\left.0, \frac{\pi}{4}\right)} =1$

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Q. The potential energy for a force field $\overrightarrow{ F }$ is given by $U(x, y)=\sin (x+y)$. The magnitude of force acting on the particle of mass $m$ at $\left(0, \frac{\pi}{4}\right)$ is :

$1$

$\sqrt{2}$

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

$0$

We use $\overrightarrow{ F } =\nabla U =-\cos (x+y) \hat{ i }-\cos (x+y) \hat{ j }$ $\left.\overrightarrow{ F }\right|_{\left(0, \frac{\pi}{4}\right)} =-\frac{1}{\sqrt{2}} \hat{ i }-\frac{1}{\sqrt{2}} \hat{ j } $ $|\overrightarrow{ F }|_{\left.0, \frac{\pi}{4}\right)} =1$

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2. The machine gun fires $240 \, $ bullets per minute. If the mass of each bullet is $10 \, g$ and the velocity of the bullets is $600 \, m \, s^{- 1}$ ,the power (in $kW$ )of the gun is

4. A particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration $a_{c}$ is varying with time $t$ as $a_{c}=k^{2} \, rt^{2}$ , where $k$ is a constant. The power delivered to the particle by the force acting on it is

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2. If $E$ and $G$ respectively denote energy and gravitational constant, then $\frac{ E }{ G }$ has the dimensions of:

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Latest Updates

We use
$\overrightarrow{ F } =\nabla U =-\cos (x+y) \hat{ i }-\cos (x+y) \hat{ j }$
$\left.\overrightarrow{ F }\right|_{\left(0, \frac{\pi}{4}\right)} =-\frac{1}{\sqrt{2}} \hat{ i }-\frac{1}{\sqrt{2}} \hat{ j } $
$|\overrightarrow{ F }|_{\left.0, \frac{\pi}{4}\right)} =1$