151972
A thick wire is stretched, so that its length become two times. Assuming that there is no change in its density, then what is the ratio of change in resistance of wire to the initial resistance of wire
1 $2: 1$
2 $4: 1$
3 $3: 1$
4 $1: 4$
Explanation:
C Resistance $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ $\mathrm{R}=\rho \frac{l \times l}{\mathrm{~A} \times l}=\rho \frac{l^{2}}{\mathrm{~V}} \quad(\because \mathrm{A} \times l=\mathrm{V})$ If density remains constant then $\therefore \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{l_{1}}{l_{2}}\right)^{2} \quad\left(\because l_{2}=2 l_{1}\right)$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{l_{1}}{2 l_{1}}\right)^{2}=\frac{1}{4}$ $\mathrm{R}_{2} =4 \mathrm{R}_{1}$ Change in resistance $=4 R_{1}-R_{1}=3 R_{1}$ $\therefore \quad \frac{\text { Changein resistance }}{\text { Original resistance }}=\frac{3 \mathrm{R}_{1}}{\mathrm{R}_{1}}=\frac{3}{1}$
MHT-CET 2004
Current Electricity
151973
The length of the resistance wire is increased by $10 \%$. What is the corresponding change in the resistance of wire?
1 $10 \%$
2 $25 \%$
3 $21 \%$
4 $9 \%$
Explanation:
C Let initial length of wire $=l$ Final length of wire $=1.1 \mathrm{l}$ Volume will remains same $l_1 \mathrm{~A}_1=l_2 \mathrm{~A}_2$ $l \mathrm{~A}_1=1.1 \mathrm{~L} \mathrm{~A}$ $\therefore \quad \mathrm{A}_2 =\frac{\mathrm{A}_1}{1.1}$ $\frac{\mathrm{R}_1}{\mathrm{R}_2} =\frac{l}{l_2} \times \frac{\mathrm{A}_2}{\mathrm{~A}_1}$ $=\frac{l}{1.1 l} \times \frac{\mathrm{A}_1 / 1.1}{\mathrm{~A}_1}$ $\frac{\mathrm{R}_1}{\mathrm{R}_2} =\frac{1}{1.21}$ $\mathrm{R}_2 =1.21 \mathrm{R}_1$ $\therefore \quad \% \text { change in resistance }$ $= \frac{\mathrm{R}_2-\mathrm{R}_1}{\mathrm{R}_1} \times 100$ $= \frac{\mathrm{R}_1 1.21-\mathrm{R}_1}{\mathrm{R}_1} \times 100=0.21 \times 100=21 \%$ $\therefore \quad \%$ change in resistance Therefore change in resistance of wire is $21 \%$
MHT-CET 2004
Current Electricity
151974
A current of $0.01 \mathrm{~mA}$ passes through the potentiometer wire of a resistivity of $10^{9} \Omega \mathrm{cm}$ and area of cross-section $10^{-2} \mathbf{c m}^{2}$. The potential gradient is
151972
A thick wire is stretched, so that its length become two times. Assuming that there is no change in its density, then what is the ratio of change in resistance of wire to the initial resistance of wire
1 $2: 1$
2 $4: 1$
3 $3: 1$
4 $1: 4$
Explanation:
C Resistance $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ $\mathrm{R}=\rho \frac{l \times l}{\mathrm{~A} \times l}=\rho \frac{l^{2}}{\mathrm{~V}} \quad(\because \mathrm{A} \times l=\mathrm{V})$ If density remains constant then $\therefore \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{l_{1}}{l_{2}}\right)^{2} \quad\left(\because l_{2}=2 l_{1}\right)$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{l_{1}}{2 l_{1}}\right)^{2}=\frac{1}{4}$ $\mathrm{R}_{2} =4 \mathrm{R}_{1}$ Change in resistance $=4 R_{1}-R_{1}=3 R_{1}$ $\therefore \quad \frac{\text { Changein resistance }}{\text { Original resistance }}=\frac{3 \mathrm{R}_{1}}{\mathrm{R}_{1}}=\frac{3}{1}$
MHT-CET 2004
Current Electricity
151973
The length of the resistance wire is increased by $10 \%$. What is the corresponding change in the resistance of wire?
1 $10 \%$
2 $25 \%$
3 $21 \%$
4 $9 \%$
Explanation:
C Let initial length of wire $=l$ Final length of wire $=1.1 \mathrm{l}$ Volume will remains same $l_1 \mathrm{~A}_1=l_2 \mathrm{~A}_2$ $l \mathrm{~A}_1=1.1 \mathrm{~L} \mathrm{~A}$ $\therefore \quad \mathrm{A}_2 =\frac{\mathrm{A}_1}{1.1}$ $\frac{\mathrm{R}_1}{\mathrm{R}_2} =\frac{l}{l_2} \times \frac{\mathrm{A}_2}{\mathrm{~A}_1}$ $=\frac{l}{1.1 l} \times \frac{\mathrm{A}_1 / 1.1}{\mathrm{~A}_1}$ $\frac{\mathrm{R}_1}{\mathrm{R}_2} =\frac{1}{1.21}$ $\mathrm{R}_2 =1.21 \mathrm{R}_1$ $\therefore \quad \% \text { change in resistance }$ $= \frac{\mathrm{R}_2-\mathrm{R}_1}{\mathrm{R}_1} \times 100$ $= \frac{\mathrm{R}_1 1.21-\mathrm{R}_1}{\mathrm{R}_1} \times 100=0.21 \times 100=21 \%$ $\therefore \quad \%$ change in resistance Therefore change in resistance of wire is $21 \%$
MHT-CET 2004
Current Electricity
151974
A current of $0.01 \mathrm{~mA}$ passes through the potentiometer wire of a resistivity of $10^{9} \Omega \mathrm{cm}$ and area of cross-section $10^{-2} \mathbf{c m}^{2}$. The potential gradient is
151972
A thick wire is stretched, so that its length become two times. Assuming that there is no change in its density, then what is the ratio of change in resistance of wire to the initial resistance of wire
1 $2: 1$
2 $4: 1$
3 $3: 1$
4 $1: 4$
Explanation:
C Resistance $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ $\mathrm{R}=\rho \frac{l \times l}{\mathrm{~A} \times l}=\rho \frac{l^{2}}{\mathrm{~V}} \quad(\because \mathrm{A} \times l=\mathrm{V})$ If density remains constant then $\therefore \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{l_{1}}{l_{2}}\right)^{2} \quad\left(\because l_{2}=2 l_{1}\right)$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{l_{1}}{2 l_{1}}\right)^{2}=\frac{1}{4}$ $\mathrm{R}_{2} =4 \mathrm{R}_{1}$ Change in resistance $=4 R_{1}-R_{1}=3 R_{1}$ $\therefore \quad \frac{\text { Changein resistance }}{\text { Original resistance }}=\frac{3 \mathrm{R}_{1}}{\mathrm{R}_{1}}=\frac{3}{1}$
MHT-CET 2004
Current Electricity
151973
The length of the resistance wire is increased by $10 \%$. What is the corresponding change in the resistance of wire?
1 $10 \%$
2 $25 \%$
3 $21 \%$
4 $9 \%$
Explanation:
C Let initial length of wire $=l$ Final length of wire $=1.1 \mathrm{l}$ Volume will remains same $l_1 \mathrm{~A}_1=l_2 \mathrm{~A}_2$ $l \mathrm{~A}_1=1.1 \mathrm{~L} \mathrm{~A}$ $\therefore \quad \mathrm{A}_2 =\frac{\mathrm{A}_1}{1.1}$ $\frac{\mathrm{R}_1}{\mathrm{R}_2} =\frac{l}{l_2} \times \frac{\mathrm{A}_2}{\mathrm{~A}_1}$ $=\frac{l}{1.1 l} \times \frac{\mathrm{A}_1 / 1.1}{\mathrm{~A}_1}$ $\frac{\mathrm{R}_1}{\mathrm{R}_2} =\frac{1}{1.21}$ $\mathrm{R}_2 =1.21 \mathrm{R}_1$ $\therefore \quad \% \text { change in resistance }$ $= \frac{\mathrm{R}_2-\mathrm{R}_1}{\mathrm{R}_1} \times 100$ $= \frac{\mathrm{R}_1 1.21-\mathrm{R}_1}{\mathrm{R}_1} \times 100=0.21 \times 100=21 \%$ $\therefore \quad \%$ change in resistance Therefore change in resistance of wire is $21 \%$
MHT-CET 2004
Current Electricity
151974
A current of $0.01 \mathrm{~mA}$ passes through the potentiometer wire of a resistivity of $10^{9} \Omega \mathrm{cm}$ and area of cross-section $10^{-2} \mathbf{c m}^{2}$. The potential gradient is
151972
A thick wire is stretched, so that its length become two times. Assuming that there is no change in its density, then what is the ratio of change in resistance of wire to the initial resistance of wire
1 $2: 1$
2 $4: 1$
3 $3: 1$
4 $1: 4$
Explanation:
C Resistance $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ $\mathrm{R}=\rho \frac{l \times l}{\mathrm{~A} \times l}=\rho \frac{l^{2}}{\mathrm{~V}} \quad(\because \mathrm{A} \times l=\mathrm{V})$ If density remains constant then $\therefore \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{l_{1}}{l_{2}}\right)^{2} \quad\left(\because l_{2}=2 l_{1}\right)$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{l_{1}}{2 l_{1}}\right)^{2}=\frac{1}{4}$ $\mathrm{R}_{2} =4 \mathrm{R}_{1}$ Change in resistance $=4 R_{1}-R_{1}=3 R_{1}$ $\therefore \quad \frac{\text { Changein resistance }}{\text { Original resistance }}=\frac{3 \mathrm{R}_{1}}{\mathrm{R}_{1}}=\frac{3}{1}$
MHT-CET 2004
Current Electricity
151973
The length of the resistance wire is increased by $10 \%$. What is the corresponding change in the resistance of wire?
1 $10 \%$
2 $25 \%$
3 $21 \%$
4 $9 \%$
Explanation:
C Let initial length of wire $=l$ Final length of wire $=1.1 \mathrm{l}$ Volume will remains same $l_1 \mathrm{~A}_1=l_2 \mathrm{~A}_2$ $l \mathrm{~A}_1=1.1 \mathrm{~L} \mathrm{~A}$ $\therefore \quad \mathrm{A}_2 =\frac{\mathrm{A}_1}{1.1}$ $\frac{\mathrm{R}_1}{\mathrm{R}_2} =\frac{l}{l_2} \times \frac{\mathrm{A}_2}{\mathrm{~A}_1}$ $=\frac{l}{1.1 l} \times \frac{\mathrm{A}_1 / 1.1}{\mathrm{~A}_1}$ $\frac{\mathrm{R}_1}{\mathrm{R}_2} =\frac{1}{1.21}$ $\mathrm{R}_2 =1.21 \mathrm{R}_1$ $\therefore \quad \% \text { change in resistance }$ $= \frac{\mathrm{R}_2-\mathrm{R}_1}{\mathrm{R}_1} \times 100$ $= \frac{\mathrm{R}_1 1.21-\mathrm{R}_1}{\mathrm{R}_1} \times 100=0.21 \times 100=21 \%$ $\therefore \quad \%$ change in resistance Therefore change in resistance of wire is $21 \%$
MHT-CET 2004
Current Electricity
151974
A current of $0.01 \mathrm{~mA}$ passes through the potentiometer wire of a resistivity of $10^{9} \Omega \mathrm{cm}$ and area of cross-section $10^{-2} \mathbf{c m}^{2}$. The potential gradient is