Physics Test - 8

According to the question

Temperature coefficient of resistance of the wire,

\(\alpha={0.002}{^{\circ}C}\)

Resistance at temperature,

\(30^{\circ} C=R_{30}=10 \Omega\)

We know,

\(R_{T}=R_{0}(1+\alpha T)\)

Putting the value of \(R_{30}, \alpha\) and \(\mathrm{T}\) in above equation we have

\(10=R_{0}(1+30 \alpha) \ldots\) (i)

After \(10 \%\) increase in resistance the new resistance becomes

\(10+10 \times \frac{10}{100}=11 \Omega\)

Again, for the new resistance we have

\(11=R_{0}(1+\alpha T) \ldots\) (ii)

Dividing equation (ii) by (i), we get

\(\frac{11}{10}=\frac{R_{0}(1+\alpha T)}{R_{0}(1+30 \alpha)}\)

\(R_{0}\) gets cancelled out as it's the same material and same reference temperature.

Or, \(\frac{11}{10}=\frac{(1+\alpha T)}{(1+30 \alpha)}\)

Or, \(11(1+30 \alpha)=10(1+\alpha T)\)

Or, \(11+330 \alpha=10+10 \alpha T\)

Or, \(1+330 \alpha=10 \alpha T\)

Or, \(\frac{1+330 \alpha}{10 \alpha}=T\)

Putting the value of \(\alpha\), we get

\(T=\frac{1+330 \times 0.002}{10 \times 0.002}\)

Or, \(T=83^{\circ} C\)

In a spherical shell, the internal energy per unit volume is given by \(\frac{U}{V} \alpha T^{4}\)

\(\Rightarrow U=C V T^{4}\)

Here \(C\) is a constant

Next the value of \(P\),

\(=\frac{1}{3}\left(\frac{U}{V}\right)=\frac{1}{3}\left(\frac{C V T^{4}}{V}\right)\)

From adiabatic expansion, \(d Q=0\) and \(d U=-d W\)

\(d(C V T)^{4}=-P d V\)

We get,

\(\Rightarrow 4 V d T=-\frac{4}{3} T d V \)

\(\Rightarrow \frac{d T}{T}=\frac{d V}{3 V}\)

On integrating,

\(\Rightarrow T V^{ \frac{1}{3}}=C^{1} \)

\(\Rightarrow T\left(\frac{4}{3} \pi R^{3}\right)^{\frac{1}{3}}=C^{1}\)

\(T R=\) Constant

\(\Rightarrow T \alpha \frac{1}{R}\)

Let conductivity of steel \(K_{\text {steel }}=k\) then from question

Conductivity of copper \(K_{\text {copper }}=9 k\)

\( \theta_{\text {copper }}=100^{\circ} C \)

\( \theta_{\text {steel }}=0^{\circ} C \)

\( l_{\text {steel }}=l_{\text {copper }}=\frac{L}{2}\)

From formula temperature of junction;

\(\theta =\frac{K_{\text {copper }} \theta_{\text {copper }} l_{\text {steel }}+K_{\text {steel }} \theta_{\text {steel }} l_{\text {copper }}}{K_{\text {copper }} l_{\text {steel }}+K_{\text {steel }} l_{\text {copper }}} \)

\( =\frac{9 k \times 100 \times \frac{L}{2}+k \times 0 \times \frac{L}{2}}{9 k \times \frac{L}{2}+k \times \frac{L}{2}} \)

\( =\frac{\frac{900}{2} k L}{\frac{10 k L}{2}}=90^{\circ} C\)

Given that the threshold frequency for metal surface corresponds to an energy \((\nu)=6.2 {eV}\)

Stopping potential for the radiation incident on it is, \(V_{0}=5 {~V}\)

We know that in the photoelectric experiment

\(E=W+K\)

\(h \nu=h \nu_{o}+e V_{o}\)

Put the given values in above equation,

\(h \nu=6.2 e V+5 e V\)

\(=11.2 e V\)

We know that,

\(h \nu=\frac{h c}{\lambda}\)

\(\Rightarrow\frac{h c}{\lambda}=11.2 e V\)

\(\Rightarrow c=3×10^{8}m/s\)

\(\Rightarrow\frac{1242 e V~ n m}{\lambda}=11.2 e V\)

\(\Rightarrow\lambda=\frac{1242 e V~ n m}{11.2 e V}\)

\(=110.89~ n m\)

\(\simeq 1100 A^{\circ}\)

Which is under the ultraviolet region.

Given,

The refracting angle of the prism = \(90^{\circ}\)

For passing the ray from prism,

\(\mu<\operatorname{cosec} \frac{A}{2}\)

\(\Rightarrow \mu<\operatorname{cosec}\left(\frac{90^{\circ}}{2}\right)\)

\(\Rightarrow \mu<\sqrt{2}\)

\(\Rightarrow \mu_{\max }=\sqrt{2}\)

Finally, the maximum refractive index of a prism \(=\sqrt{2}\)

A time period of a satellite is the time taken by the satellite to go once around the earth.

\(T=2 \pi \sqrt{\frac{r^{3}}{M G}}\),

Where \(r=R+D\),

if \(\mathrm{D}\) is very small wrt to \(R\) so \(\mathrm{D}\) can be neglected.

So, \(T=2 \pi \sqrt{\frac{1}{\mathrm{M} \sigma}}\),

where \(\sigma\) is density of the planet.

So, here clearly we can see if the orbit of the satellite is very close to the planet then its period depends only on the density of the planet.

As we know,

Angular momentum \(\mathrm{J}=\mathrm{m \omega r}^{2}....(1)\)

where \(\omega\) is the angular velocity, \(r\) is the radius ofthe orbit.

As we know,

Magnetic moment \(\mu=i \mathrm{~A}\)

\(=\frac{\mathrm{e} \omega}{2 \pi} \pi r^{2}.....(2)\)

By dividing equation (1) and (2), we get

\(\therefore \frac{\mu}{\mathrm{J}}=\frac{\frac{e \omega}{2 \pi} \pi r^{2}}{m \omega r^{2}}=\frac{e}{2 m}\)

\(\Rightarrow {\mu}=\frac{\mathrm{eJ}}{2 \mathrm{~m}}\)

Total energy \(=-3.4 \mathrm{ev}\)

In Bohr's model of H atom

\(\therefore \quad\) K.E. \(=|\mathrm{TE}|=\frac{|\mathrm{U}|}{2}\)

K.E. \(=-(T . E .)=3.4 \mathrm{eV}\)

\(P . E .=2(T . E)=2 \times(-3.4 \mathrm{eV})=-6.8 \mathrm{eV}\)

As we know,

Work done by a variable force is given by,

\(W=\int_{a}^{b} F(x) d x\)

Let \(x\) be the distance covered.

Given that \(F \propto \frac{1}{x}\)

Work done, \((W)=\int_{a}^{b} F(x) d x\)

\(\therefore W \propto \int_{a}^{b} \frac{1}{x} d x \)

\(\Rightarrow W \propto[\ln (x)]_{a}^{b} \)

\(\Rightarrow W \propto \ln (b)-\ln (a) \)

\(\Rightarrow W \propto \ln \left(\frac{b}{a}\right)\)