152557
A cell has an emf $1.5 \mathrm{~V}$. When connected across an external resistance of $2 \Omega$, the terminal potential difference falls to $1.0 \mathrm{~V}$. The internal resistance of the cell is
1 $2 \Omega$
2 $1.5 \Omega$
3 $1.0 \Omega$
4 $0.5 \Omega$
Explanation:
C Given that, $\mathrm{E}=1.5 \mathrm{~V}$ $\mathrm{V}=1 \text { Volt }$ $\mathrm{R}=2 \Omega$ Then internal resistance $r=\left(\frac{E}{V}-1\right) R$ $r=\left(\frac{1.5}{1}-1\right) \times 2$ $r=(1.5-1) \times 2$ $r=0.5 \times 2=1 \Omega$ $r=1 \Omega$
AIPMT-2000
Current Electricity
152558
For a cell, the terminal potential difference is $2.2 \mathrm{~V}$ when circuit is open and reduces to $1.8 \mathrm{~V}$ when cell is connected to a resistance $R=5 \Omega$, the internal resistance $(r)$ of cell is
1 $\frac{10}{9} \Omega$
2 $\frac{9}{10} \Omega$
3 $\frac{11}{9} \Omega$
4 $\frac{5}{9} \Omega$
Explanation:
A Given that, $\mathrm{E}=2.2 \mathrm{~V}$ $\mathrm{V}=1.8 \text { volt }$ $\mathrm{R}=5 \Omega$ $\mathrm{r}=\left(\frac{\mathrm{E}}{\mathrm{V}}-1\right) \mathrm{R}$ $\mathrm{r}=\left(\frac{2.2}{1.8}-1\right) \times 5$ $r=\left(\frac{2.2-1.8}{1.8}\right) \times 5$ $r=\frac{0.4}{0.8} \times 5$ $r=\frac{2.0}{1.8}$ $r=\frac{10}{9} \Omega$
AIPMT-2002
Current Electricity
152559
A $6 \mathrm{~V}$ battery is connected to the terminals of a $3 \mathrm{~m}$ long wire of uniform thickness and resistance of $100 \Omega$. The difference of potential between two points on the wire separated by a distance of $50 \mathrm{~cm}$ will be
1 $2 \mathrm{~V}$
2 $3 \mathrm{~V}$
3 $1 \mathrm{~V}$
4 $1.5 \mathrm{~V}$
Explanation:
C Given that, The potential drop across $3 \mathrm{~m}$ will be $6 \mathrm{~V}$. then potential drop at $50 \mathrm{~cm}$ will be. $l_{1}=3 \mathrm{~m} \mathrm{E}_{1}=6 \text { volt }$ $l_{2}=0.5 \mathrm{~m} \mathrm{E}_{2}=?$ $\frac{\mathrm{E}_{1}}{l_{1}}=\frac{\mathrm{E}_{2}}{l_{2}}$ So, $\quad \frac{6}{3}=\frac{\mathrm{E}_{2}}{0.5}$ $\mathrm{E}_{2}=2 \times 0.5$ $\mathrm{E}_{2}=1 \mathrm{Volt}$
AIPMT-2004
Current Electricity
152560
A student measures the terminal potential difference ( $V$ ) of a cell (of emf $E$ and internal resistance $r$ ) as a function of the current (I) flowing through it. The slope and intercept of the graph between $V$ and $I$, respectively, equal to
1 E and - r
2 $-\mathrm{r}$ and $\mathrm{E}$
3 $\mathrm{r}$ and $-\mathrm{E}$
4 - E and $r$
Explanation:
B Then potential difference $(\mathrm{V})=\mathrm{E}-$ ir Comparing equation $(\mathrm{y})=\mathrm{mx}+\mathrm{C}$ So Intercept $=\mathrm{E}$ And slope $=-r$
AIPMT-2009
Current Electricity
152562
A wire of a certain material is stretched slowly by 10 percent. Its new resistance and specific resistance become respectively
1 1.2 times, 1.1 times
2 1.21 times, same
3 Both remain the same
4 1.1 times, 1.1 times
Explanation:
B We know that Resistance $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ But when length increase $10 \%$ therefore $l^{\prime}=l+\frac{l}{10}=\frac{11 l}{10}$ and therefore area become $\mathrm{A}^{\prime}=\frac{10 \mathrm{~A}}{11}$ then new resistance $\mathrm{R}^{\prime}=\frac{\rho\left(\frac{11 l}{10}\right)}{\frac{10 \mathrm{~A}}{11}}=\frac{\rho l}{\mathrm{~A}} \times \frac{(11)^{2}}{(10)^{2}}$ therefore $\mathrm{R}^{\prime}=1.21 \mathrm{R}$ therefore new resistance increase 1.21 time but specific resistance remains some.
152557
A cell has an emf $1.5 \mathrm{~V}$. When connected across an external resistance of $2 \Omega$, the terminal potential difference falls to $1.0 \mathrm{~V}$. The internal resistance of the cell is
1 $2 \Omega$
2 $1.5 \Omega$
3 $1.0 \Omega$
4 $0.5 \Omega$
Explanation:
C Given that, $\mathrm{E}=1.5 \mathrm{~V}$ $\mathrm{V}=1 \text { Volt }$ $\mathrm{R}=2 \Omega$ Then internal resistance $r=\left(\frac{E}{V}-1\right) R$ $r=\left(\frac{1.5}{1}-1\right) \times 2$ $r=(1.5-1) \times 2$ $r=0.5 \times 2=1 \Omega$ $r=1 \Omega$
AIPMT-2000
Current Electricity
152558
For a cell, the terminal potential difference is $2.2 \mathrm{~V}$ when circuit is open and reduces to $1.8 \mathrm{~V}$ when cell is connected to a resistance $R=5 \Omega$, the internal resistance $(r)$ of cell is
1 $\frac{10}{9} \Omega$
2 $\frac{9}{10} \Omega$
3 $\frac{11}{9} \Omega$
4 $\frac{5}{9} \Omega$
Explanation:
A Given that, $\mathrm{E}=2.2 \mathrm{~V}$ $\mathrm{V}=1.8 \text { volt }$ $\mathrm{R}=5 \Omega$ $\mathrm{r}=\left(\frac{\mathrm{E}}{\mathrm{V}}-1\right) \mathrm{R}$ $\mathrm{r}=\left(\frac{2.2}{1.8}-1\right) \times 5$ $r=\left(\frac{2.2-1.8}{1.8}\right) \times 5$ $r=\frac{0.4}{0.8} \times 5$ $r=\frac{2.0}{1.8}$ $r=\frac{10}{9} \Omega$
AIPMT-2002
Current Electricity
152559
A $6 \mathrm{~V}$ battery is connected to the terminals of a $3 \mathrm{~m}$ long wire of uniform thickness and resistance of $100 \Omega$. The difference of potential between two points on the wire separated by a distance of $50 \mathrm{~cm}$ will be
1 $2 \mathrm{~V}$
2 $3 \mathrm{~V}$
3 $1 \mathrm{~V}$
4 $1.5 \mathrm{~V}$
Explanation:
C Given that, The potential drop across $3 \mathrm{~m}$ will be $6 \mathrm{~V}$. then potential drop at $50 \mathrm{~cm}$ will be. $l_{1}=3 \mathrm{~m} \mathrm{E}_{1}=6 \text { volt }$ $l_{2}=0.5 \mathrm{~m} \mathrm{E}_{2}=?$ $\frac{\mathrm{E}_{1}}{l_{1}}=\frac{\mathrm{E}_{2}}{l_{2}}$ So, $\quad \frac{6}{3}=\frac{\mathrm{E}_{2}}{0.5}$ $\mathrm{E}_{2}=2 \times 0.5$ $\mathrm{E}_{2}=1 \mathrm{Volt}$
AIPMT-2004
Current Electricity
152560
A student measures the terminal potential difference ( $V$ ) of a cell (of emf $E$ and internal resistance $r$ ) as a function of the current (I) flowing through it. The slope and intercept of the graph between $V$ and $I$, respectively, equal to
1 E and - r
2 $-\mathrm{r}$ and $\mathrm{E}$
3 $\mathrm{r}$ and $-\mathrm{E}$
4 - E and $r$
Explanation:
B Then potential difference $(\mathrm{V})=\mathrm{E}-$ ir Comparing equation $(\mathrm{y})=\mathrm{mx}+\mathrm{C}$ So Intercept $=\mathrm{E}$ And slope $=-r$
AIPMT-2009
Current Electricity
152562
A wire of a certain material is stretched slowly by 10 percent. Its new resistance and specific resistance become respectively
1 1.2 times, 1.1 times
2 1.21 times, same
3 Both remain the same
4 1.1 times, 1.1 times
Explanation:
B We know that Resistance $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ But when length increase $10 \%$ therefore $l^{\prime}=l+\frac{l}{10}=\frac{11 l}{10}$ and therefore area become $\mathrm{A}^{\prime}=\frac{10 \mathrm{~A}}{11}$ then new resistance $\mathrm{R}^{\prime}=\frac{\rho\left(\frac{11 l}{10}\right)}{\frac{10 \mathrm{~A}}{11}}=\frac{\rho l}{\mathrm{~A}} \times \frac{(11)^{2}}{(10)^{2}}$ therefore $\mathrm{R}^{\prime}=1.21 \mathrm{R}$ therefore new resistance increase 1.21 time but specific resistance remains some.
152557
A cell has an emf $1.5 \mathrm{~V}$. When connected across an external resistance of $2 \Omega$, the terminal potential difference falls to $1.0 \mathrm{~V}$. The internal resistance of the cell is
1 $2 \Omega$
2 $1.5 \Omega$
3 $1.0 \Omega$
4 $0.5 \Omega$
Explanation:
C Given that, $\mathrm{E}=1.5 \mathrm{~V}$ $\mathrm{V}=1 \text { Volt }$ $\mathrm{R}=2 \Omega$ Then internal resistance $r=\left(\frac{E}{V}-1\right) R$ $r=\left(\frac{1.5}{1}-1\right) \times 2$ $r=(1.5-1) \times 2$ $r=0.5 \times 2=1 \Omega$ $r=1 \Omega$
AIPMT-2000
Current Electricity
152558
For a cell, the terminal potential difference is $2.2 \mathrm{~V}$ when circuit is open and reduces to $1.8 \mathrm{~V}$ when cell is connected to a resistance $R=5 \Omega$, the internal resistance $(r)$ of cell is
1 $\frac{10}{9} \Omega$
2 $\frac{9}{10} \Omega$
3 $\frac{11}{9} \Omega$
4 $\frac{5}{9} \Omega$
Explanation:
A Given that, $\mathrm{E}=2.2 \mathrm{~V}$ $\mathrm{V}=1.8 \text { volt }$ $\mathrm{R}=5 \Omega$ $\mathrm{r}=\left(\frac{\mathrm{E}}{\mathrm{V}}-1\right) \mathrm{R}$ $\mathrm{r}=\left(\frac{2.2}{1.8}-1\right) \times 5$ $r=\left(\frac{2.2-1.8}{1.8}\right) \times 5$ $r=\frac{0.4}{0.8} \times 5$ $r=\frac{2.0}{1.8}$ $r=\frac{10}{9} \Omega$
AIPMT-2002
Current Electricity
152559
A $6 \mathrm{~V}$ battery is connected to the terminals of a $3 \mathrm{~m}$ long wire of uniform thickness and resistance of $100 \Omega$. The difference of potential between two points on the wire separated by a distance of $50 \mathrm{~cm}$ will be
1 $2 \mathrm{~V}$
2 $3 \mathrm{~V}$
3 $1 \mathrm{~V}$
4 $1.5 \mathrm{~V}$
Explanation:
C Given that, The potential drop across $3 \mathrm{~m}$ will be $6 \mathrm{~V}$. then potential drop at $50 \mathrm{~cm}$ will be. $l_{1}=3 \mathrm{~m} \mathrm{E}_{1}=6 \text { volt }$ $l_{2}=0.5 \mathrm{~m} \mathrm{E}_{2}=?$ $\frac{\mathrm{E}_{1}}{l_{1}}=\frac{\mathrm{E}_{2}}{l_{2}}$ So, $\quad \frac{6}{3}=\frac{\mathrm{E}_{2}}{0.5}$ $\mathrm{E}_{2}=2 \times 0.5$ $\mathrm{E}_{2}=1 \mathrm{Volt}$
AIPMT-2004
Current Electricity
152560
A student measures the terminal potential difference ( $V$ ) of a cell (of emf $E$ and internal resistance $r$ ) as a function of the current (I) flowing through it. The slope and intercept of the graph between $V$ and $I$, respectively, equal to
1 E and - r
2 $-\mathrm{r}$ and $\mathrm{E}$
3 $\mathrm{r}$ and $-\mathrm{E}$
4 - E and $r$
Explanation:
B Then potential difference $(\mathrm{V})=\mathrm{E}-$ ir Comparing equation $(\mathrm{y})=\mathrm{mx}+\mathrm{C}$ So Intercept $=\mathrm{E}$ And slope $=-r$
AIPMT-2009
Current Electricity
152562
A wire of a certain material is stretched slowly by 10 percent. Its new resistance and specific resistance become respectively
1 1.2 times, 1.1 times
2 1.21 times, same
3 Both remain the same
4 1.1 times, 1.1 times
Explanation:
B We know that Resistance $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ But when length increase $10 \%$ therefore $l^{\prime}=l+\frac{l}{10}=\frac{11 l}{10}$ and therefore area become $\mathrm{A}^{\prime}=\frac{10 \mathrm{~A}}{11}$ then new resistance $\mathrm{R}^{\prime}=\frac{\rho\left(\frac{11 l}{10}\right)}{\frac{10 \mathrm{~A}}{11}}=\frac{\rho l}{\mathrm{~A}} \times \frac{(11)^{2}}{(10)^{2}}$ therefore $\mathrm{R}^{\prime}=1.21 \mathrm{R}$ therefore new resistance increase 1.21 time but specific resistance remains some.
152557
A cell has an emf $1.5 \mathrm{~V}$. When connected across an external resistance of $2 \Omega$, the terminal potential difference falls to $1.0 \mathrm{~V}$. The internal resistance of the cell is
1 $2 \Omega$
2 $1.5 \Omega$
3 $1.0 \Omega$
4 $0.5 \Omega$
Explanation:
C Given that, $\mathrm{E}=1.5 \mathrm{~V}$ $\mathrm{V}=1 \text { Volt }$ $\mathrm{R}=2 \Omega$ Then internal resistance $r=\left(\frac{E}{V}-1\right) R$ $r=\left(\frac{1.5}{1}-1\right) \times 2$ $r=(1.5-1) \times 2$ $r=0.5 \times 2=1 \Omega$ $r=1 \Omega$
AIPMT-2000
Current Electricity
152558
For a cell, the terminal potential difference is $2.2 \mathrm{~V}$ when circuit is open and reduces to $1.8 \mathrm{~V}$ when cell is connected to a resistance $R=5 \Omega$, the internal resistance $(r)$ of cell is
1 $\frac{10}{9} \Omega$
2 $\frac{9}{10} \Omega$
3 $\frac{11}{9} \Omega$
4 $\frac{5}{9} \Omega$
Explanation:
A Given that, $\mathrm{E}=2.2 \mathrm{~V}$ $\mathrm{V}=1.8 \text { volt }$ $\mathrm{R}=5 \Omega$ $\mathrm{r}=\left(\frac{\mathrm{E}}{\mathrm{V}}-1\right) \mathrm{R}$ $\mathrm{r}=\left(\frac{2.2}{1.8}-1\right) \times 5$ $r=\left(\frac{2.2-1.8}{1.8}\right) \times 5$ $r=\frac{0.4}{0.8} \times 5$ $r=\frac{2.0}{1.8}$ $r=\frac{10}{9} \Omega$
AIPMT-2002
Current Electricity
152559
A $6 \mathrm{~V}$ battery is connected to the terminals of a $3 \mathrm{~m}$ long wire of uniform thickness and resistance of $100 \Omega$. The difference of potential between two points on the wire separated by a distance of $50 \mathrm{~cm}$ will be
1 $2 \mathrm{~V}$
2 $3 \mathrm{~V}$
3 $1 \mathrm{~V}$
4 $1.5 \mathrm{~V}$
Explanation:
C Given that, The potential drop across $3 \mathrm{~m}$ will be $6 \mathrm{~V}$. then potential drop at $50 \mathrm{~cm}$ will be. $l_{1}=3 \mathrm{~m} \mathrm{E}_{1}=6 \text { volt }$ $l_{2}=0.5 \mathrm{~m} \mathrm{E}_{2}=?$ $\frac{\mathrm{E}_{1}}{l_{1}}=\frac{\mathrm{E}_{2}}{l_{2}}$ So, $\quad \frac{6}{3}=\frac{\mathrm{E}_{2}}{0.5}$ $\mathrm{E}_{2}=2 \times 0.5$ $\mathrm{E}_{2}=1 \mathrm{Volt}$
AIPMT-2004
Current Electricity
152560
A student measures the terminal potential difference ( $V$ ) of a cell (of emf $E$ and internal resistance $r$ ) as a function of the current (I) flowing through it. The slope and intercept of the graph between $V$ and $I$, respectively, equal to
1 E and - r
2 $-\mathrm{r}$ and $\mathrm{E}$
3 $\mathrm{r}$ and $-\mathrm{E}$
4 - E and $r$
Explanation:
B Then potential difference $(\mathrm{V})=\mathrm{E}-$ ir Comparing equation $(\mathrm{y})=\mathrm{mx}+\mathrm{C}$ So Intercept $=\mathrm{E}$ And slope $=-r$
AIPMT-2009
Current Electricity
152562
A wire of a certain material is stretched slowly by 10 percent. Its new resistance and specific resistance become respectively
1 1.2 times, 1.1 times
2 1.21 times, same
3 Both remain the same
4 1.1 times, 1.1 times
Explanation:
B We know that Resistance $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ But when length increase $10 \%$ therefore $l^{\prime}=l+\frac{l}{10}=\frac{11 l}{10}$ and therefore area become $\mathrm{A}^{\prime}=\frac{10 \mathrm{~A}}{11}$ then new resistance $\mathrm{R}^{\prime}=\frac{\rho\left(\frac{11 l}{10}\right)}{\frac{10 \mathrm{~A}}{11}}=\frac{\rho l}{\mathrm{~A}} \times \frac{(11)^{2}}{(10)^{2}}$ therefore $\mathrm{R}^{\prime}=1.21 \mathrm{R}$ therefore new resistance increase 1.21 time but specific resistance remains some.
152557
A cell has an emf $1.5 \mathrm{~V}$. When connected across an external resistance of $2 \Omega$, the terminal potential difference falls to $1.0 \mathrm{~V}$. The internal resistance of the cell is
1 $2 \Omega$
2 $1.5 \Omega$
3 $1.0 \Omega$
4 $0.5 \Omega$
Explanation:
C Given that, $\mathrm{E}=1.5 \mathrm{~V}$ $\mathrm{V}=1 \text { Volt }$ $\mathrm{R}=2 \Omega$ Then internal resistance $r=\left(\frac{E}{V}-1\right) R$ $r=\left(\frac{1.5}{1}-1\right) \times 2$ $r=(1.5-1) \times 2$ $r=0.5 \times 2=1 \Omega$ $r=1 \Omega$
AIPMT-2000
Current Electricity
152558
For a cell, the terminal potential difference is $2.2 \mathrm{~V}$ when circuit is open and reduces to $1.8 \mathrm{~V}$ when cell is connected to a resistance $R=5 \Omega$, the internal resistance $(r)$ of cell is
1 $\frac{10}{9} \Omega$
2 $\frac{9}{10} \Omega$
3 $\frac{11}{9} \Omega$
4 $\frac{5}{9} \Omega$
Explanation:
A Given that, $\mathrm{E}=2.2 \mathrm{~V}$ $\mathrm{V}=1.8 \text { volt }$ $\mathrm{R}=5 \Omega$ $\mathrm{r}=\left(\frac{\mathrm{E}}{\mathrm{V}}-1\right) \mathrm{R}$ $\mathrm{r}=\left(\frac{2.2}{1.8}-1\right) \times 5$ $r=\left(\frac{2.2-1.8}{1.8}\right) \times 5$ $r=\frac{0.4}{0.8} \times 5$ $r=\frac{2.0}{1.8}$ $r=\frac{10}{9} \Omega$
AIPMT-2002
Current Electricity
152559
A $6 \mathrm{~V}$ battery is connected to the terminals of a $3 \mathrm{~m}$ long wire of uniform thickness and resistance of $100 \Omega$. The difference of potential between two points on the wire separated by a distance of $50 \mathrm{~cm}$ will be
1 $2 \mathrm{~V}$
2 $3 \mathrm{~V}$
3 $1 \mathrm{~V}$
4 $1.5 \mathrm{~V}$
Explanation:
C Given that, The potential drop across $3 \mathrm{~m}$ will be $6 \mathrm{~V}$. then potential drop at $50 \mathrm{~cm}$ will be. $l_{1}=3 \mathrm{~m} \mathrm{E}_{1}=6 \text { volt }$ $l_{2}=0.5 \mathrm{~m} \mathrm{E}_{2}=?$ $\frac{\mathrm{E}_{1}}{l_{1}}=\frac{\mathrm{E}_{2}}{l_{2}}$ So, $\quad \frac{6}{3}=\frac{\mathrm{E}_{2}}{0.5}$ $\mathrm{E}_{2}=2 \times 0.5$ $\mathrm{E}_{2}=1 \mathrm{Volt}$
AIPMT-2004
Current Electricity
152560
A student measures the terminal potential difference ( $V$ ) of a cell (of emf $E$ and internal resistance $r$ ) as a function of the current (I) flowing through it. The slope and intercept of the graph between $V$ and $I$, respectively, equal to
1 E and - r
2 $-\mathrm{r}$ and $\mathrm{E}$
3 $\mathrm{r}$ and $-\mathrm{E}$
4 - E and $r$
Explanation:
B Then potential difference $(\mathrm{V})=\mathrm{E}-$ ir Comparing equation $(\mathrm{y})=\mathrm{mx}+\mathrm{C}$ So Intercept $=\mathrm{E}$ And slope $=-r$
AIPMT-2009
Current Electricity
152562
A wire of a certain material is stretched slowly by 10 percent. Its new resistance and specific resistance become respectively
1 1.2 times, 1.1 times
2 1.21 times, same
3 Both remain the same
4 1.1 times, 1.1 times
Explanation:
B We know that Resistance $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}$ But when length increase $10 \%$ therefore $l^{\prime}=l+\frac{l}{10}=\frac{11 l}{10}$ and therefore area become $\mathrm{A}^{\prime}=\frac{10 \mathrm{~A}}{11}$ then new resistance $\mathrm{R}^{\prime}=\frac{\rho\left(\frac{11 l}{10}\right)}{\frac{10 \mathrm{~A}}{11}}=\frac{\rho l}{\mathrm{~A}} \times \frac{(11)^{2}}{(10)^{2}}$ therefore $\mathrm{R}^{\prime}=1.21 \mathrm{R}$ therefore new resistance increase 1.21 time but specific resistance remains some.
