D According to question, power of two electric bulb $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$. We know that, Power, $(\mathrm{P})=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ So, $\quad \mathrm{P}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{1}} \Rightarrow \mathrm{R}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{P}_{1}}$ Similarly, $\mathrm{P}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{2}} \Rightarrow \mathrm{R}_{2}=\frac{\mathrm{V}^{2}}{\mathrm{P}_{2}}$ These two bulb connected in series their resistance will be in series and power consumption, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$ $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{~V}^{2}\left(\frac{1}{\mathrm{P}_{1}}+\frac{1}{\mathrm{P}_{2}}\right)}$ $\mathrm{P}=\frac{1}{\frac{\mathrm{P}_{1}+\mathrm{P}_{2}}{\mathrm{P}_{1} \mathrm{P}_{2}}}$ $\mathrm{P}=\frac{\mathrm{P}_{1} \mathrm{P}_{2}}{\mathrm{P}_{1}+\mathrm{P}_{2}}$
Tripura-27.04.2022
Current Electricity
152646
A direct current of $4 \mathrm{~A}$ and an alternating current of peak value flow through resistance of $3 \Omega$ and $2 \Omega$ respectively. The ratio of heat produced in the two resistances in same interval of time will be:
1 $3: 2$
2 $3: 1$
3 $3: 4$
4 $4: 3$
Explanation:
B Given, \(\text {Current }(i)=4 \mathrm{~A}\) \(\mathrm{R}_1=3 \Omega, \mathrm{R}_2=2 \Omega\) We know that, Heat produced $(\mathrm{H})=\mathrm{i}^{2} \mathrm{Rt}$ For DC supply, $\mathrm{H}_{1}=\mathrm{i}_{1}^{2} \mathrm{R}_{1} \mathrm{t}$ $\mathrm{H}_{1}=(4)^{2} \times 3 \times \mathrm{t}$ $\mathrm{H}_{1}=48 \mathrm{t}$ For AC supply, $\therefore \mathrm{H}_{2} =\mathrm{i}_{\mathrm{rms}}^{2} \mathrm{R}_{2} \mathrm{t}$ $\mathrm{H}_{2} =\left(\frac{\mathrm{i}_{\mathrm{o}}}{\sqrt{2}}\right)^{2} \mathrm{R}_{2} \mathrm{t} \quad\left[\because \mathrm{i}_{\mathrm{rms}}=\frac{\mathrm{i}_{\mathrm{o}}}{\sqrt{2}}\right]$ $\mathrm{H}_{2} =\left(\frac{4}{\sqrt{2}}\right)^{2} \times 2 \mathrm{t}$ $\mathrm{H}_{2}=16 \mathrm{t}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{48 \mathrm{t}}{16 \mathrm{t}}=\frac{3}{1}$ $\mathrm{H}_{1}: \mathrm{H}_{2}=3: 1$ The ratio of heat produced in the two resistances in same interval of time will be $3: 1$.
JEE Main-27.07.2022
Current Electricity
152647
A bulb of power $660 \mathrm{~W}$ radiates uniformly in all directions. The pressure exerted by the radiation on the surface at a distance of $5 \mathrm{~m}$ is
1 $5 \times 10^{-8} \mathrm{~Pa}$
2 $2 \times 10^{-9} \mathrm{~Pa}$
3 $7 \times 10^{-9} \mathrm{~Pa}$
4 $\frac{3}{\pi} \times 10^{-8} \mathrm{~Pa}$
Explanation:
C Given, Power, \(\mathrm{P}=660 \mathrm{~W}\) Distance \(=5 \mathrm{~m}\) By using formula- \(\text { Pressure of radiation, } \mathrm{P}_{\mathrm{r}}=\frac{\mathrm{P}}{4 \pi \mathrm{r}^2 \mathrm{C}}\) \(\mathrm{P}_{\mathrm{r}} =\frac{660}{4 \times 3.14 \times 25 \times 3 \times 10^8 \mathrm{~m}}\) \(=7 \times 10^{-9} \mathrm{~Pa}\)
AP EAMCET-07.07.2022
Current Electricity
152648 In the above circuit, the heat produced in $5 \Omega$ resistance is 10 calories per second. The heat produced in $4 \Omega$ resistance is
1 $1 \mathrm{cal} / \mathrm{sec}$.
2 $2 \mathrm{cal} / \mathrm{sec}$.
3 $3 \mathrm{cal} / \mathrm{sec}$.
4 $4 \mathrm{cal} / \mathrm{sec}$.
Explanation:
B We know that - $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}$ $\frac{\mathrm{H}}{\mathrm{t}}=\mathrm{I}^{2} \mathrm{R}$ For $5 \Omega$ $(\mathrm{H} / \mathrm{t}) =10$ $\therefore \text { From }(\mathrm{i})-$ $10 =\mathrm{I}_{1}^{2} \times 5$ $\mathrm{I}_{1} =\sqrt{2} \mathrm{~A}$ $(\mathrm{H} / \mathrm{t})=10 \mathrm{cal} / \mathrm{sec}$ Now, $\mathrm{V}_{1}=\mathrm{V}_{2}$ $\mathrm{I}_{1} \mathrm{R}_{1}=\mathrm{I}_{2} \mathrm{R}_{2}$ $5 \mathrm{I}_{1}=\mathrm{I}_{2}(4+6)$ $5 \mathrm{I}_{1}=10 \mathrm{I}_{2}$ $\mathrm{I}_{2}=\frac{\mathrm{I}_{1}}{2}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} \mathrm{~A}$ Hence, heat in $4 \Omega$ resistance per second - $(\mathrm{H} / \mathrm{t}) =\mathrm{I}_{2}^{2} \mathrm{R}$ $=\left(\frac{1}{\sqrt{2}}\right)^{2} \times 4$ $=2 \mathrm{cal} / \text { second }$
D According to question, power of two electric bulb $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$. We know that, Power, $(\mathrm{P})=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ So, $\quad \mathrm{P}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{1}} \Rightarrow \mathrm{R}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{P}_{1}}$ Similarly, $\mathrm{P}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{2}} \Rightarrow \mathrm{R}_{2}=\frac{\mathrm{V}^{2}}{\mathrm{P}_{2}}$ These two bulb connected in series their resistance will be in series and power consumption, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$ $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{~V}^{2}\left(\frac{1}{\mathrm{P}_{1}}+\frac{1}{\mathrm{P}_{2}}\right)}$ $\mathrm{P}=\frac{1}{\frac{\mathrm{P}_{1}+\mathrm{P}_{2}}{\mathrm{P}_{1} \mathrm{P}_{2}}}$ $\mathrm{P}=\frac{\mathrm{P}_{1} \mathrm{P}_{2}}{\mathrm{P}_{1}+\mathrm{P}_{2}}$
Tripura-27.04.2022
Current Electricity
152646
A direct current of $4 \mathrm{~A}$ and an alternating current of peak value flow through resistance of $3 \Omega$ and $2 \Omega$ respectively. The ratio of heat produced in the two resistances in same interval of time will be:
1 $3: 2$
2 $3: 1$
3 $3: 4$
4 $4: 3$
Explanation:
B Given, \(\text {Current }(i)=4 \mathrm{~A}\) \(\mathrm{R}_1=3 \Omega, \mathrm{R}_2=2 \Omega\) We know that, Heat produced $(\mathrm{H})=\mathrm{i}^{2} \mathrm{Rt}$ For DC supply, $\mathrm{H}_{1}=\mathrm{i}_{1}^{2} \mathrm{R}_{1} \mathrm{t}$ $\mathrm{H}_{1}=(4)^{2} \times 3 \times \mathrm{t}$ $\mathrm{H}_{1}=48 \mathrm{t}$ For AC supply, $\therefore \mathrm{H}_{2} =\mathrm{i}_{\mathrm{rms}}^{2} \mathrm{R}_{2} \mathrm{t}$ $\mathrm{H}_{2} =\left(\frac{\mathrm{i}_{\mathrm{o}}}{\sqrt{2}}\right)^{2} \mathrm{R}_{2} \mathrm{t} \quad\left[\because \mathrm{i}_{\mathrm{rms}}=\frac{\mathrm{i}_{\mathrm{o}}}{\sqrt{2}}\right]$ $\mathrm{H}_{2} =\left(\frac{4}{\sqrt{2}}\right)^{2} \times 2 \mathrm{t}$ $\mathrm{H}_{2}=16 \mathrm{t}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{48 \mathrm{t}}{16 \mathrm{t}}=\frac{3}{1}$ $\mathrm{H}_{1}: \mathrm{H}_{2}=3: 1$ The ratio of heat produced in the two resistances in same interval of time will be $3: 1$.
JEE Main-27.07.2022
Current Electricity
152647
A bulb of power $660 \mathrm{~W}$ radiates uniformly in all directions. The pressure exerted by the radiation on the surface at a distance of $5 \mathrm{~m}$ is
1 $5 \times 10^{-8} \mathrm{~Pa}$
2 $2 \times 10^{-9} \mathrm{~Pa}$
3 $7 \times 10^{-9} \mathrm{~Pa}$
4 $\frac{3}{\pi} \times 10^{-8} \mathrm{~Pa}$
Explanation:
C Given, Power, \(\mathrm{P}=660 \mathrm{~W}\) Distance \(=5 \mathrm{~m}\) By using formula- \(\text { Pressure of radiation, } \mathrm{P}_{\mathrm{r}}=\frac{\mathrm{P}}{4 \pi \mathrm{r}^2 \mathrm{C}}\) \(\mathrm{P}_{\mathrm{r}} =\frac{660}{4 \times 3.14 \times 25 \times 3 \times 10^8 \mathrm{~m}}\) \(=7 \times 10^{-9} \mathrm{~Pa}\)
AP EAMCET-07.07.2022
Current Electricity
152648 In the above circuit, the heat produced in $5 \Omega$ resistance is 10 calories per second. The heat produced in $4 \Omega$ resistance is
1 $1 \mathrm{cal} / \mathrm{sec}$.
2 $2 \mathrm{cal} / \mathrm{sec}$.
3 $3 \mathrm{cal} / \mathrm{sec}$.
4 $4 \mathrm{cal} / \mathrm{sec}$.
Explanation:
B We know that - $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}$ $\frac{\mathrm{H}}{\mathrm{t}}=\mathrm{I}^{2} \mathrm{R}$ For $5 \Omega$ $(\mathrm{H} / \mathrm{t}) =10$ $\therefore \text { From }(\mathrm{i})-$ $10 =\mathrm{I}_{1}^{2} \times 5$ $\mathrm{I}_{1} =\sqrt{2} \mathrm{~A}$ $(\mathrm{H} / \mathrm{t})=10 \mathrm{cal} / \mathrm{sec}$ Now, $\mathrm{V}_{1}=\mathrm{V}_{2}$ $\mathrm{I}_{1} \mathrm{R}_{1}=\mathrm{I}_{2} \mathrm{R}_{2}$ $5 \mathrm{I}_{1}=\mathrm{I}_{2}(4+6)$ $5 \mathrm{I}_{1}=10 \mathrm{I}_{2}$ $\mathrm{I}_{2}=\frac{\mathrm{I}_{1}}{2}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} \mathrm{~A}$ Hence, heat in $4 \Omega$ resistance per second - $(\mathrm{H} / \mathrm{t}) =\mathrm{I}_{2}^{2} \mathrm{R}$ $=\left(\frac{1}{\sqrt{2}}\right)^{2} \times 4$ $=2 \mathrm{cal} / \text { second }$
D According to question, power of two electric bulb $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$. We know that, Power, $(\mathrm{P})=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ So, $\quad \mathrm{P}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{1}} \Rightarrow \mathrm{R}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{P}_{1}}$ Similarly, $\mathrm{P}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{2}} \Rightarrow \mathrm{R}_{2}=\frac{\mathrm{V}^{2}}{\mathrm{P}_{2}}$ These two bulb connected in series their resistance will be in series and power consumption, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$ $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{~V}^{2}\left(\frac{1}{\mathrm{P}_{1}}+\frac{1}{\mathrm{P}_{2}}\right)}$ $\mathrm{P}=\frac{1}{\frac{\mathrm{P}_{1}+\mathrm{P}_{2}}{\mathrm{P}_{1} \mathrm{P}_{2}}}$ $\mathrm{P}=\frac{\mathrm{P}_{1} \mathrm{P}_{2}}{\mathrm{P}_{1}+\mathrm{P}_{2}}$
Tripura-27.04.2022
Current Electricity
152646
A direct current of $4 \mathrm{~A}$ and an alternating current of peak value flow through resistance of $3 \Omega$ and $2 \Omega$ respectively. The ratio of heat produced in the two resistances in same interval of time will be:
1 $3: 2$
2 $3: 1$
3 $3: 4$
4 $4: 3$
Explanation:
B Given, \(\text {Current }(i)=4 \mathrm{~A}\) \(\mathrm{R}_1=3 \Omega, \mathrm{R}_2=2 \Omega\) We know that, Heat produced $(\mathrm{H})=\mathrm{i}^{2} \mathrm{Rt}$ For DC supply, $\mathrm{H}_{1}=\mathrm{i}_{1}^{2} \mathrm{R}_{1} \mathrm{t}$ $\mathrm{H}_{1}=(4)^{2} \times 3 \times \mathrm{t}$ $\mathrm{H}_{1}=48 \mathrm{t}$ For AC supply, $\therefore \mathrm{H}_{2} =\mathrm{i}_{\mathrm{rms}}^{2} \mathrm{R}_{2} \mathrm{t}$ $\mathrm{H}_{2} =\left(\frac{\mathrm{i}_{\mathrm{o}}}{\sqrt{2}}\right)^{2} \mathrm{R}_{2} \mathrm{t} \quad\left[\because \mathrm{i}_{\mathrm{rms}}=\frac{\mathrm{i}_{\mathrm{o}}}{\sqrt{2}}\right]$ $\mathrm{H}_{2} =\left(\frac{4}{\sqrt{2}}\right)^{2} \times 2 \mathrm{t}$ $\mathrm{H}_{2}=16 \mathrm{t}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{48 \mathrm{t}}{16 \mathrm{t}}=\frac{3}{1}$ $\mathrm{H}_{1}: \mathrm{H}_{2}=3: 1$ The ratio of heat produced in the two resistances in same interval of time will be $3: 1$.
JEE Main-27.07.2022
Current Electricity
152647
A bulb of power $660 \mathrm{~W}$ radiates uniformly in all directions. The pressure exerted by the radiation on the surface at a distance of $5 \mathrm{~m}$ is
1 $5 \times 10^{-8} \mathrm{~Pa}$
2 $2 \times 10^{-9} \mathrm{~Pa}$
3 $7 \times 10^{-9} \mathrm{~Pa}$
4 $\frac{3}{\pi} \times 10^{-8} \mathrm{~Pa}$
Explanation:
C Given, Power, \(\mathrm{P}=660 \mathrm{~W}\) Distance \(=5 \mathrm{~m}\) By using formula- \(\text { Pressure of radiation, } \mathrm{P}_{\mathrm{r}}=\frac{\mathrm{P}}{4 \pi \mathrm{r}^2 \mathrm{C}}\) \(\mathrm{P}_{\mathrm{r}} =\frac{660}{4 \times 3.14 \times 25 \times 3 \times 10^8 \mathrm{~m}}\) \(=7 \times 10^{-9} \mathrm{~Pa}\)
AP EAMCET-07.07.2022
Current Electricity
152648 In the above circuit, the heat produced in $5 \Omega$ resistance is 10 calories per second. The heat produced in $4 \Omega$ resistance is
1 $1 \mathrm{cal} / \mathrm{sec}$.
2 $2 \mathrm{cal} / \mathrm{sec}$.
3 $3 \mathrm{cal} / \mathrm{sec}$.
4 $4 \mathrm{cal} / \mathrm{sec}$.
Explanation:
B We know that - $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}$ $\frac{\mathrm{H}}{\mathrm{t}}=\mathrm{I}^{2} \mathrm{R}$ For $5 \Omega$ $(\mathrm{H} / \mathrm{t}) =10$ $\therefore \text { From }(\mathrm{i})-$ $10 =\mathrm{I}_{1}^{2} \times 5$ $\mathrm{I}_{1} =\sqrt{2} \mathrm{~A}$ $(\mathrm{H} / \mathrm{t})=10 \mathrm{cal} / \mathrm{sec}$ Now, $\mathrm{V}_{1}=\mathrm{V}_{2}$ $\mathrm{I}_{1} \mathrm{R}_{1}=\mathrm{I}_{2} \mathrm{R}_{2}$ $5 \mathrm{I}_{1}=\mathrm{I}_{2}(4+6)$ $5 \mathrm{I}_{1}=10 \mathrm{I}_{2}$ $\mathrm{I}_{2}=\frac{\mathrm{I}_{1}}{2}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} \mathrm{~A}$ Hence, heat in $4 \Omega$ resistance per second - $(\mathrm{H} / \mathrm{t}) =\mathrm{I}_{2}^{2} \mathrm{R}$ $=\left(\frac{1}{\sqrt{2}}\right)^{2} \times 4$ $=2 \mathrm{cal} / \text { second }$
D According to question, power of two electric bulb $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$. We know that, Power, $(\mathrm{P})=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ So, $\quad \mathrm{P}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{1}} \Rightarrow \mathrm{R}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{P}_{1}}$ Similarly, $\mathrm{P}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{2}} \Rightarrow \mathrm{R}_{2}=\frac{\mathrm{V}^{2}}{\mathrm{P}_{2}}$ These two bulb connected in series their resistance will be in series and power consumption, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$ $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{~V}^{2}\left(\frac{1}{\mathrm{P}_{1}}+\frac{1}{\mathrm{P}_{2}}\right)}$ $\mathrm{P}=\frac{1}{\frac{\mathrm{P}_{1}+\mathrm{P}_{2}}{\mathrm{P}_{1} \mathrm{P}_{2}}}$ $\mathrm{P}=\frac{\mathrm{P}_{1} \mathrm{P}_{2}}{\mathrm{P}_{1}+\mathrm{P}_{2}}$
Tripura-27.04.2022
Current Electricity
152646
A direct current of $4 \mathrm{~A}$ and an alternating current of peak value flow through resistance of $3 \Omega$ and $2 \Omega$ respectively. The ratio of heat produced in the two resistances in same interval of time will be:
1 $3: 2$
2 $3: 1$
3 $3: 4$
4 $4: 3$
Explanation:
B Given, \(\text {Current }(i)=4 \mathrm{~A}\) \(\mathrm{R}_1=3 \Omega, \mathrm{R}_2=2 \Omega\) We know that, Heat produced $(\mathrm{H})=\mathrm{i}^{2} \mathrm{Rt}$ For DC supply, $\mathrm{H}_{1}=\mathrm{i}_{1}^{2} \mathrm{R}_{1} \mathrm{t}$ $\mathrm{H}_{1}=(4)^{2} \times 3 \times \mathrm{t}$ $\mathrm{H}_{1}=48 \mathrm{t}$ For AC supply, $\therefore \mathrm{H}_{2} =\mathrm{i}_{\mathrm{rms}}^{2} \mathrm{R}_{2} \mathrm{t}$ $\mathrm{H}_{2} =\left(\frac{\mathrm{i}_{\mathrm{o}}}{\sqrt{2}}\right)^{2} \mathrm{R}_{2} \mathrm{t} \quad\left[\because \mathrm{i}_{\mathrm{rms}}=\frac{\mathrm{i}_{\mathrm{o}}}{\sqrt{2}}\right]$ $\mathrm{H}_{2} =\left(\frac{4}{\sqrt{2}}\right)^{2} \times 2 \mathrm{t}$ $\mathrm{H}_{2}=16 \mathrm{t}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{48 \mathrm{t}}{16 \mathrm{t}}=\frac{3}{1}$ $\mathrm{H}_{1}: \mathrm{H}_{2}=3: 1$ The ratio of heat produced in the two resistances in same interval of time will be $3: 1$.
JEE Main-27.07.2022
Current Electricity
152647
A bulb of power $660 \mathrm{~W}$ radiates uniformly in all directions. The pressure exerted by the radiation on the surface at a distance of $5 \mathrm{~m}$ is
1 $5 \times 10^{-8} \mathrm{~Pa}$
2 $2 \times 10^{-9} \mathrm{~Pa}$
3 $7 \times 10^{-9} \mathrm{~Pa}$
4 $\frac{3}{\pi} \times 10^{-8} \mathrm{~Pa}$
Explanation:
C Given, Power, \(\mathrm{P}=660 \mathrm{~W}\) Distance \(=5 \mathrm{~m}\) By using formula- \(\text { Pressure of radiation, } \mathrm{P}_{\mathrm{r}}=\frac{\mathrm{P}}{4 \pi \mathrm{r}^2 \mathrm{C}}\) \(\mathrm{P}_{\mathrm{r}} =\frac{660}{4 \times 3.14 \times 25 \times 3 \times 10^8 \mathrm{~m}}\) \(=7 \times 10^{-9} \mathrm{~Pa}\)
AP EAMCET-07.07.2022
Current Electricity
152648 In the above circuit, the heat produced in $5 \Omega$ resistance is 10 calories per second. The heat produced in $4 \Omega$ resistance is
1 $1 \mathrm{cal} / \mathrm{sec}$.
2 $2 \mathrm{cal} / \mathrm{sec}$.
3 $3 \mathrm{cal} / \mathrm{sec}$.
4 $4 \mathrm{cal} / \mathrm{sec}$.
Explanation:
B We know that - $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}$ $\frac{\mathrm{H}}{\mathrm{t}}=\mathrm{I}^{2} \mathrm{R}$ For $5 \Omega$ $(\mathrm{H} / \mathrm{t}) =10$ $\therefore \text { From }(\mathrm{i})-$ $10 =\mathrm{I}_{1}^{2} \times 5$ $\mathrm{I}_{1} =\sqrt{2} \mathrm{~A}$ $(\mathrm{H} / \mathrm{t})=10 \mathrm{cal} / \mathrm{sec}$ Now, $\mathrm{V}_{1}=\mathrm{V}_{2}$ $\mathrm{I}_{1} \mathrm{R}_{1}=\mathrm{I}_{2} \mathrm{R}_{2}$ $5 \mathrm{I}_{1}=\mathrm{I}_{2}(4+6)$ $5 \mathrm{I}_{1}=10 \mathrm{I}_{2}$ $\mathrm{I}_{2}=\frac{\mathrm{I}_{1}}{2}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} \mathrm{~A}$ Hence, heat in $4 \Omega$ resistance per second - $(\mathrm{H} / \mathrm{t}) =\mathrm{I}_{2}^{2} \mathrm{R}$ $=\left(\frac{1}{\sqrt{2}}\right)^{2} \times 4$ $=2 \mathrm{cal} / \text { second }$
