Physics

Hint: Potential energy is maximum at the highest point.

Step 1: Find the time taken to reach the maximum height.
\(v=u+at\\ 0=10-10t\\ t=1~\text{s}\)
Therefore, the potential energy will be maximum at \(t=1~\text{s}\)$$

Step 2: Find height where kinetic energy is equal to potential energy
\(v^2=u^2-2gh \dots(1)\\ K.E=P.E\\ \frac{1}{2}mv^2=mgh\\ \text{From equation (1)}\\ u^2-2gh=2gh\\ \Rightarrow u^2=4gh\\ \Rightarrow h=\frac{u^2}{4g}\\ \Rightarrow h= 2.5~\text{m} ~~~~\left[u= 10~\text{m/s}, g = 10 ~\text{m/s}^2\right]\)
Therefore, kinetic energy is equal to the potential energy at a height of \(2.5~\text{m}\) from the ground.
Hence, option (3) is the correct answer.

Given, 
Mass of the planet = ${\mathrm{M}}_{\mathrm{p}}$
The diameter of the planet = ${\mathrm{D}}_{\mathrm{p}}$
Mass of the freely falling particle = m 
Force, 
$\mathrm{F}=\frac{{\mathrm{GM}}_{\mathrm{p}}\mathrm{m}}{{{\mathrm{R}}_{\mathrm{p}}}^2}$ 
Here, $\mathrm{F}=\frac{{\mathrm{GM}}_{\mathrm{p}}\mathrm{m}}{{({\mathrm{D}}_{\mathrm{p}}/2)}^2}$ 
$\text{F=}\frac{4{\mathrm{GM}}_{\mathrm{p}}\mathrm{m}}{{\mathrm{D}}_{\mathrm{p}}^2}$
Acceleration due to gravity, 
$\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{4{\mathrm{GM}}_{\mathrm{p}}}{{\mathrm{D}}_{\mathrm{p}}^2}$

Hint: \(P=\dfrac{V_{\text {total }}^2}{R_{\text {total }}}\)

Step: Find the power loss in the wire.
The total potential drop \((V_{total})\) is given by;
\(\Rightarrow V_{\text {total }}=\text { Potential drop per km } \times \text { Distance in km} \)
\(\Rightarrow V_{\text {total }}=8 ~\text{V}/ \text{km} \times 150 ~\text{km}=1200~\text{V}\)
The total resistance \((R_{total})\) over the distance is given by;
\(\Rightarrow R_{\text {total }}=\text { Resistance per km} \times \text { Distance in km} \)
\(\Rightarrow R_{\text {total }}=0.5~ \Omega / \text{km} \times 150 ~\text{km}=75 ~\Omega \)
The power loss \((P)\) in the wire is given by;
\(\Rightarrow P=\frac{V_{\text {total }}^2}{R_{\text {total }}}\)
\(\Rightarrow P=\frac{(1200 ~\text{V})^2}{75~\Omega} \Rightarrow P=19.2~\text{kW}\)
Hence, option (2) is the correct answer.

Hint: \(K.E_{\max }=h_f-\phi \)

Step: Find the velocity of the electron ejected from a silver surface by the ultraviolet light.
Given,
\( \lambda_0=3250 \times 10^{-10} \mathrm{~m}, ~~\lambda=2536 \times 10^{-10} \mathrm{~m} \)
\(\phi=\frac{{hc}}{\lambda_0}=\frac{4.14 \times 10^{-15} \times 3 \times 10^8}{3250 \times 10^{-10}}=3.82 ~\mathrm{eV} \)
\({h} \nu=\frac{{hc}}{\lambda}=\frac{4.14 \times 10^{-15} \times 3 \times 10^8}{2536 \times 10^{-10}}=4.89~ \mathrm{eV}\)
According to Einstein’s photoelectric equation,
\(K.E_{\max }=h_f-\phi \)
\(K.E_{\max }=(4.89-3.82) \mathrm{eV}=1.077 \mathrm{eV}\)
\(\frac{1}{2} m v^2=1.077 \times 1.6 \times 10^{-19} \)
\(v=\sqrt{\frac{2 \times 1.077 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}} \)
\(\Rightarrow v=0.6 \times 10^6{\mathrm{~m}}/{\mathrm{~s}}\)
Hence, option (1) is the correct answer.