152790
Balancing point of a potentiometer shifts from a length of $60 \mathrm{~cm}$ to $40 \mathrm{~cm}$ by shunting the cell with a $4 \mathrm{ohm}$ resistance. What is internal resistance of the cell?
1 $1 \Omega$
2 $2 \Omega$
3 $4 \Omega$
4 $6 \Omega$
Explanation:
B Given, $\mathrm{L}_{1}=60 \mathrm{~cm}, \quad \mathrm{~L}_{2}=40 \mathrm{~cm}, \quad \mathrm{R}=4 \Omega$ The internal resistance $r$ of a cell is given by $r=R\left[\frac{L_{1}}{\mathrm{~L}_{2}}-1\right]$ $=4\left[\frac{60}{40}-1\right]=4\left[\frac{3}{2}-1\right]$ $=4 \times \frac{1}{2}=2 \Omega$
TS EAMCET 18.07.2022
Current Electricity
152792
A unit positive charge has to be brought from infinity to a midpoint between two charges 20 $\mu \mathrm{C}$ and $10 \mu \mathrm{C}$ separated by a distance of $50 \mathrm{~m}$. How much work will be required?
152793
A radio transmitter operates at a frequency $880 \mathrm{kHz}$ and a power of $10 \mathrm{~kW}$. What is the number of photons emitted per second?
1 $1.50 \times 10^{25}$
2 $1.60 \times 10^{30}$
3 $1.72 \times 10^{31}$
4 $2.80 \times 10^{30}$
Explanation:
C P $=10 \mathrm{~kW}$ $=10 \times 10^{3} \mathrm{~W}$ $v=880 \mathrm{KHz}$ $v=880 \times 10^{3} \mathrm{~Hz}$ Number of Photons emitted per second $\mathrm{N}=\frac{\mathrm{P}}{\mathrm{hv}}$ $=\frac{10 \times 10^{3}}{6.6 \times 10^{-34} \times 880 \times 10^{3}}$ $=1.72 \times 10^{31} \text { Photons/second }$
AP EAMCET-03.09.2021
Current Electricity
152795
In a potentiometer experiment the balancing with a cell is at length $250 \mathrm{~cm}$. In shunting the cell with a resistance of $2 \Omega$, the balancing length becomes $125 \mathrm{~cm}$. The internal resistance of the cell is
1 $2 \Omega$
2 $4 \Omega$
3 $0.5 \Omega$
4 $1 \Omega$
Explanation:
A Given, $l_{1}=250 \mathrm{~cm}$ $l_{2}=125 \mathrm{~cm}$ $\mathrm{R}=2 \Omega$ Internal resistance of a cell $\mathrm{r}$ is given by $\mathrm{r}=\mathrm{R}\left[\frac{l_{1}}{l_{2}}-1\right]$ $\mathrm{r}=2\left[\frac{250}{125}-1\right]$ $\mathrm{r}=2[2-1]$ $\mathrm{r}=2 \times 1$ $\mathrm{r}=2 \Omega$
152790
Balancing point of a potentiometer shifts from a length of $60 \mathrm{~cm}$ to $40 \mathrm{~cm}$ by shunting the cell with a $4 \mathrm{ohm}$ resistance. What is internal resistance of the cell?
1 $1 \Omega$
2 $2 \Omega$
3 $4 \Omega$
4 $6 \Omega$
Explanation:
B Given, $\mathrm{L}_{1}=60 \mathrm{~cm}, \quad \mathrm{~L}_{2}=40 \mathrm{~cm}, \quad \mathrm{R}=4 \Omega$ The internal resistance $r$ of a cell is given by $r=R\left[\frac{L_{1}}{\mathrm{~L}_{2}}-1\right]$ $=4\left[\frac{60}{40}-1\right]=4\left[\frac{3}{2}-1\right]$ $=4 \times \frac{1}{2}=2 \Omega$
TS EAMCET 18.07.2022
Current Electricity
152792
A unit positive charge has to be brought from infinity to a midpoint between two charges 20 $\mu \mathrm{C}$ and $10 \mu \mathrm{C}$ separated by a distance of $50 \mathrm{~m}$. How much work will be required?
152793
A radio transmitter operates at a frequency $880 \mathrm{kHz}$ and a power of $10 \mathrm{~kW}$. What is the number of photons emitted per second?
1 $1.50 \times 10^{25}$
2 $1.60 \times 10^{30}$
3 $1.72 \times 10^{31}$
4 $2.80 \times 10^{30}$
Explanation:
C P $=10 \mathrm{~kW}$ $=10 \times 10^{3} \mathrm{~W}$ $v=880 \mathrm{KHz}$ $v=880 \times 10^{3} \mathrm{~Hz}$ Number of Photons emitted per second $\mathrm{N}=\frac{\mathrm{P}}{\mathrm{hv}}$ $=\frac{10 \times 10^{3}}{6.6 \times 10^{-34} \times 880 \times 10^{3}}$ $=1.72 \times 10^{31} \text { Photons/second }$
AP EAMCET-03.09.2021
Current Electricity
152795
In a potentiometer experiment the balancing with a cell is at length $250 \mathrm{~cm}$. In shunting the cell with a resistance of $2 \Omega$, the balancing length becomes $125 \mathrm{~cm}$. The internal resistance of the cell is
1 $2 \Omega$
2 $4 \Omega$
3 $0.5 \Omega$
4 $1 \Omega$
Explanation:
A Given, $l_{1}=250 \mathrm{~cm}$ $l_{2}=125 \mathrm{~cm}$ $\mathrm{R}=2 \Omega$ Internal resistance of a cell $\mathrm{r}$ is given by $\mathrm{r}=\mathrm{R}\left[\frac{l_{1}}{l_{2}}-1\right]$ $\mathrm{r}=2\left[\frac{250}{125}-1\right]$ $\mathrm{r}=2[2-1]$ $\mathrm{r}=2 \times 1$ $\mathrm{r}=2 \Omega$
152790
Balancing point of a potentiometer shifts from a length of $60 \mathrm{~cm}$ to $40 \mathrm{~cm}$ by shunting the cell with a $4 \mathrm{ohm}$ resistance. What is internal resistance of the cell?
1 $1 \Omega$
2 $2 \Omega$
3 $4 \Omega$
4 $6 \Omega$
Explanation:
B Given, $\mathrm{L}_{1}=60 \mathrm{~cm}, \quad \mathrm{~L}_{2}=40 \mathrm{~cm}, \quad \mathrm{R}=4 \Omega$ The internal resistance $r$ of a cell is given by $r=R\left[\frac{L_{1}}{\mathrm{~L}_{2}}-1\right]$ $=4\left[\frac{60}{40}-1\right]=4\left[\frac{3}{2}-1\right]$ $=4 \times \frac{1}{2}=2 \Omega$
TS EAMCET 18.07.2022
Current Electricity
152792
A unit positive charge has to be brought from infinity to a midpoint between two charges 20 $\mu \mathrm{C}$ and $10 \mu \mathrm{C}$ separated by a distance of $50 \mathrm{~m}$. How much work will be required?
152793
A radio transmitter operates at a frequency $880 \mathrm{kHz}$ and a power of $10 \mathrm{~kW}$. What is the number of photons emitted per second?
1 $1.50 \times 10^{25}$
2 $1.60 \times 10^{30}$
3 $1.72 \times 10^{31}$
4 $2.80 \times 10^{30}$
Explanation:
C P $=10 \mathrm{~kW}$ $=10 \times 10^{3} \mathrm{~W}$ $v=880 \mathrm{KHz}$ $v=880 \times 10^{3} \mathrm{~Hz}$ Number of Photons emitted per second $\mathrm{N}=\frac{\mathrm{P}}{\mathrm{hv}}$ $=\frac{10 \times 10^{3}}{6.6 \times 10^{-34} \times 880 \times 10^{3}}$ $=1.72 \times 10^{31} \text { Photons/second }$
AP EAMCET-03.09.2021
Current Electricity
152795
In a potentiometer experiment the balancing with a cell is at length $250 \mathrm{~cm}$. In shunting the cell with a resistance of $2 \Omega$, the balancing length becomes $125 \mathrm{~cm}$. The internal resistance of the cell is
1 $2 \Omega$
2 $4 \Omega$
3 $0.5 \Omega$
4 $1 \Omega$
Explanation:
A Given, $l_{1}=250 \mathrm{~cm}$ $l_{2}=125 \mathrm{~cm}$ $\mathrm{R}=2 \Omega$ Internal resistance of a cell $\mathrm{r}$ is given by $\mathrm{r}=\mathrm{R}\left[\frac{l_{1}}{l_{2}}-1\right]$ $\mathrm{r}=2\left[\frac{250}{125}-1\right]$ $\mathrm{r}=2[2-1]$ $\mathrm{r}=2 \times 1$ $\mathrm{r}=2 \Omega$
152790
Balancing point of a potentiometer shifts from a length of $60 \mathrm{~cm}$ to $40 \mathrm{~cm}$ by shunting the cell with a $4 \mathrm{ohm}$ resistance. What is internal resistance of the cell?
1 $1 \Omega$
2 $2 \Omega$
3 $4 \Omega$
4 $6 \Omega$
Explanation:
B Given, $\mathrm{L}_{1}=60 \mathrm{~cm}, \quad \mathrm{~L}_{2}=40 \mathrm{~cm}, \quad \mathrm{R}=4 \Omega$ The internal resistance $r$ of a cell is given by $r=R\left[\frac{L_{1}}{\mathrm{~L}_{2}}-1\right]$ $=4\left[\frac{60}{40}-1\right]=4\left[\frac{3}{2}-1\right]$ $=4 \times \frac{1}{2}=2 \Omega$
TS EAMCET 18.07.2022
Current Electricity
152792
A unit positive charge has to be brought from infinity to a midpoint between two charges 20 $\mu \mathrm{C}$ and $10 \mu \mathrm{C}$ separated by a distance of $50 \mathrm{~m}$. How much work will be required?
152793
A radio transmitter operates at a frequency $880 \mathrm{kHz}$ and a power of $10 \mathrm{~kW}$. What is the number of photons emitted per second?
1 $1.50 \times 10^{25}$
2 $1.60 \times 10^{30}$
3 $1.72 \times 10^{31}$
4 $2.80 \times 10^{30}$
Explanation:
C P $=10 \mathrm{~kW}$ $=10 \times 10^{3} \mathrm{~W}$ $v=880 \mathrm{KHz}$ $v=880 \times 10^{3} \mathrm{~Hz}$ Number of Photons emitted per second $\mathrm{N}=\frac{\mathrm{P}}{\mathrm{hv}}$ $=\frac{10 \times 10^{3}}{6.6 \times 10^{-34} \times 880 \times 10^{3}}$ $=1.72 \times 10^{31} \text { Photons/second }$
AP EAMCET-03.09.2021
Current Electricity
152795
In a potentiometer experiment the balancing with a cell is at length $250 \mathrm{~cm}$. In shunting the cell with a resistance of $2 \Omega$, the balancing length becomes $125 \mathrm{~cm}$. The internal resistance of the cell is
1 $2 \Omega$
2 $4 \Omega$
3 $0.5 \Omega$
4 $1 \Omega$
Explanation:
A Given, $l_{1}=250 \mathrm{~cm}$ $l_{2}=125 \mathrm{~cm}$ $\mathrm{R}=2 \Omega$ Internal resistance of a cell $\mathrm{r}$ is given by $\mathrm{r}=\mathrm{R}\left[\frac{l_{1}}{l_{2}}-1\right]$ $\mathrm{r}=2\left[\frac{250}{125}-1\right]$ $\mathrm{r}=2[2-1]$ $\mathrm{r}=2 \times 1$ $\mathrm{r}=2 \Omega$