Physics Test - 7

Given,

Molar heat capacity of a gas at constant volume \(\mathrm{C_{V}} = 8 \mathrm{cal} / \mathrm{mol}-\mathrm{K}\), Gas constant \(\mathrm{R}=4 \mathrm{cal} / \mathrm{mol}-\mathrm{K}\).

For an ideal gas,

\(\Rightarrow \mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}\) (Meyer's equation)

\(\Rightarrow\mathrm{C}_{\mathrm{p}}=\mathrm{C}_{\mathrm{V}}+\mathrm{R}=(8+4)=12 \mathrm{cal}/\mathrm{mol}-\mathrm{K}\)

The ratio of the two principal specific heat is represented by \(\gamma\).

\(\Rightarrow\gamma=\frac{\mathrm{C}_{p}}{\mathrm{C}_{v}}\)

\(\Rightarrow\gamma=\frac{12}{8}=1.5\)

Let the charge distribution be as shown in the figure:

\(\therefore V_{A}-V_{B}=\frac{q}{C_{1}}-E+\frac{q}{C_{2}}\)

or, \(\left(V_{A}-V_{B}\right)+E=q\left(\frac{1}{C_{1}}+\frac{1}{C_{2}}\right)\)

\(\left(V_{A}-V_{B}\right)+E=\frac{q\left(C_{2}+C_{1}\right)}{C_{1} C_{2}}\)

\(\therefore q=\frac{\left[\left(V_{A}-V_{B}\right)+E\right] C_{1} C_{2}}{C_{1}+C_{2}}\)

Voltage across \(C_{1}\) is \(V_{1}=\frac{q}{C_{1}}\)

\(=\frac{\left[\left(V_{A}-V_{B}\right)+E\right] C_{2}}{C_{1}+C_{2}}\)

\(=\frac{(5+10) 2.0}{1.0+2.0}\)

\(=10\) Volt

Voltage across \(C_{2}\) is \(V_{2}=\frac{q}{C_{2}}\)

\(=\frac{\left[\left(V_{A}-V_{B}\right)+E\left[C_{1}\right.\right.}{C_{1}+C_{2}}\)

\(=\frac{(5+10) 1.0}{1.0 + 2.0}\)

\(=5\) Volt

Concept:

The basic condition of a rigid body is,

Considering two points \(P\) and \(Q\) on a rigid body.

The relative position of point \(P\) with respect to \(Q\) is \(r_{P Q}\).

The relative velocity of \(P\) with respect to \(Q\) is \(V_{P Q}=\omega \dot{r}_{P Q}\).

If the relative distance between the \(P\) and \(Q\) does not change then \(V_{P Q}=\omega r_{P Q}\) \(\mathrm{V}_{\mathrm{PQ}}=\omega \times \dot{\mathrm{r}}_\frac{\mathrm{P}}{Q}\)

\(\omega\) is the angular velocity of the rigid body.

Since \(V_{P Q}\) is the cross product of \(\omega\) and \(\dot{\mathrm{r}}_{\frac{P}Q}\) the relative velocity of \(P\) with respect to Q will be perpendicular to PQ.

Given,

A rod of length \(l\) and radius \(\mathrm{r}\) is joined to a rod of length \(\frac l2\) and radius \(\frac r2\) of same material.

As we know,

Torque, \(\mathrm{τ}=\mathrm{C} . {\theta}=\frac{\pi \eta \mathrm{r}^{4} {\theta}}{2 \mathrm{~L}}=\mathrm{constant}\)

Here, \(\pi\) and \(\eta\) are constant.

Then, according to the question,

\(\frac{\pi \eta \mathrm{r}^{4}\left(\theta-\theta_{0}\right)}{2 l}=\frac{\pi \eta(\frac{\mathrm{r}}  2)^{4}\left(\theta_{0}-\theta^{\prime}\right)}{2(\frac l 2)}\)

At fixed end, \(\theta' = 0\)

\(\Rightarrow \frac{\left({\theta}-{\theta}_{0}\right)}{2}=\frac{{\theta}_{0}}{\mathrm{1 6}}\)

\(\Rightarrow \theta_{0}=\frac{8 \theta} 9\)

lf the speed of rotation of the earth about its axis ln the weight of the body at the equator will decrease.

This is due to the gravity's force is resolved into a centripetal force that acts parallel to the equator.Thus, at the Poles, apparent weight is the same as \(mg\).

At the equator, apparent weight will be \(m g^{\prime}=\mathrm{m}\left(g-R \omega^{2}\right)\)

Where, \(\omega\) is the angular velocity.

At equilibrium, \(m u \cos \alpha=2 m v^{\prime}\)

\(\Rightarrow v^{\prime}=\frac{u \cos \alpha}{2}\)

\(\Rightarrow \frac{1}{2} =\frac{{v}^{\prime}}{u\cos \alpha}.......(1)\)

As we know that,

The coefficient of restitution,

\(e=\frac{{v}^{\prime}}{u\cos \alpha}....(2)\)

From equation (1) and (2), we get, \(e=\frac{1}{2}\)

A magnifying glass uses a convex lens because these lenses cause light rays to converge, or come together. As the light rays enter the convex lens present in the magnifying glass, these rays become focused on a specific focal point in front of the center of the lens. Furthermore, if the magnifying glass is at an optimal distance, this generates the maximum magnification of the object.

Therefore, a magnifying glass comprises a simple convex lens.

In practice, the work output of a machine is always less than the work input due to the effect of friction.

The work output of a machine is never equal to the work input because some of the work done by the machine is used to overcome the friction created by the use of the machine.

This is the reason why the efficiency of a machine can never be 100%.

The efficiency is the work output, divided by the work input, and expressed as a percentage.