NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
151975
Which statement is correct for the given circuit?
1 I through $R_{1}>$ I through $R_{2}$
2 I through $R_{3}>I$ through $R_{2}$ and $R_{1}$
3 I through $R_{2}>I$ through $R_{3}$ and $R_{1}$
4 I is same in $R_{1}, R_{2}$ and $R_{3}$
Explanation:
D In series current will always same. So, I will be same in $\mathrm{R}_{1}, \mathrm{R}_{2}, \mathrm{R}_{3}$.
VITEEE-2016
Current Electricity
151976
Dimensions of a block are $1 \mathrm{~cm} \times 1 \mathrm{~cm} \times 100 \mathrm{~cm}$. If specific resistance of its material is $3 \times 10^{-7} \Omega \mathrm{m}$, then the resistance between the opposite rectangular faces is
151977
Equal amounts of a metal are converted into cylindrical wires of different lengths $(L)$ and cross- sectional area (A). The wire with the maximum resistance is the one, which has
1 length $=\mathrm{L}$ and area $=\mathrm{A}$
2 length $=\frac{\mathrm{L}}{2}$ and area $=2 \mathrm{~A}$
3 length $=2 \mathrm{~L}$ and area $=\frac{\mathrm{A}}{2}$
4 all have the same resistance, as the amount of the metal is the same
Explanation:
C Resistance, $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ Where, $\rho$ is the resistivity, $\mathrm{R} \propto \frac{l}{\mathrm{~A}}$ According to question resistance $(\mathrm{R})$ will be maximum when- $l=2 \mathrm{~L}$ $\& \mathrm{~A}=\mathrm{A} / 2$
VITEEE-2008
Current Electricity
151978
The resistance of a field coil measures $50 \Omega$ at $20^{\circ} \mathrm{C}$ and $65 \Omega$ at $70^{\circ} \mathrm{C}$. The temperature coefficient of resistance is
1 $0.0086 /{ }^{\circ} \mathrm{C}$
2 $0.0068 /{ }^{\circ} \mathrm{C}$
3 $0.0096 /{ }^{\circ} \mathrm{C}$
4 $0.0999 /{ }^{\circ} \mathrm{C}$
Explanation:
B Given, Resistance at $20^{\circ} \mathrm{C}=50 \Omega$ Resistance at $70^{\circ} \mathrm{C}=65 \Omega$ $\because \quad \mathrm{R}=\mathrm{R}_{\mathrm{o}}(1+\alpha \mathrm{t})$ $\therefore \quad 50=\mathrm{R}_{\mathrm{o}}(1+\alpha 20)$ $65=\mathrm{R}_{\mathrm{o}}(1+\alpha 70)$ Dividing equation (ii) and equation (i), we get- $\frac{65}{50}=\frac{R_{o}(1+\alpha 70)}{R_{o}(1+\alpha 20)}$ $65+1300 \alpha=50+3500 \alpha$ $65-50=3500 \alpha-1300 \alpha$ $15=2200 \alpha$ $\alpha=\frac{15}{2200}$ $\alpha=0.0068 /{ }^{\circ} \mathrm{C}$
151975
Which statement is correct for the given circuit?
1 I through $R_{1}>$ I through $R_{2}$
2 I through $R_{3}>I$ through $R_{2}$ and $R_{1}$
3 I through $R_{2}>I$ through $R_{3}$ and $R_{1}$
4 I is same in $R_{1}, R_{2}$ and $R_{3}$
Explanation:
D In series current will always same. So, I will be same in $\mathrm{R}_{1}, \mathrm{R}_{2}, \mathrm{R}_{3}$.
VITEEE-2016
Current Electricity
151976
Dimensions of a block are $1 \mathrm{~cm} \times 1 \mathrm{~cm} \times 100 \mathrm{~cm}$. If specific resistance of its material is $3 \times 10^{-7} \Omega \mathrm{m}$, then the resistance between the opposite rectangular faces is
151977
Equal amounts of a metal are converted into cylindrical wires of different lengths $(L)$ and cross- sectional area (A). The wire with the maximum resistance is the one, which has
1 length $=\mathrm{L}$ and area $=\mathrm{A}$
2 length $=\frac{\mathrm{L}}{2}$ and area $=2 \mathrm{~A}$
3 length $=2 \mathrm{~L}$ and area $=\frac{\mathrm{A}}{2}$
4 all have the same resistance, as the amount of the metal is the same
Explanation:
C Resistance, $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ Where, $\rho$ is the resistivity, $\mathrm{R} \propto \frac{l}{\mathrm{~A}}$ According to question resistance $(\mathrm{R})$ will be maximum when- $l=2 \mathrm{~L}$ $\& \mathrm{~A}=\mathrm{A} / 2$
VITEEE-2008
Current Electricity
151978
The resistance of a field coil measures $50 \Omega$ at $20^{\circ} \mathrm{C}$ and $65 \Omega$ at $70^{\circ} \mathrm{C}$. The temperature coefficient of resistance is
1 $0.0086 /{ }^{\circ} \mathrm{C}$
2 $0.0068 /{ }^{\circ} \mathrm{C}$
3 $0.0096 /{ }^{\circ} \mathrm{C}$
4 $0.0999 /{ }^{\circ} \mathrm{C}$
Explanation:
B Given, Resistance at $20^{\circ} \mathrm{C}=50 \Omega$ Resistance at $70^{\circ} \mathrm{C}=65 \Omega$ $\because \quad \mathrm{R}=\mathrm{R}_{\mathrm{o}}(1+\alpha \mathrm{t})$ $\therefore \quad 50=\mathrm{R}_{\mathrm{o}}(1+\alpha 20)$ $65=\mathrm{R}_{\mathrm{o}}(1+\alpha 70)$ Dividing equation (ii) and equation (i), we get- $\frac{65}{50}=\frac{R_{o}(1+\alpha 70)}{R_{o}(1+\alpha 20)}$ $65+1300 \alpha=50+3500 \alpha$ $65-50=3500 \alpha-1300 \alpha$ $15=2200 \alpha$ $\alpha=\frac{15}{2200}$ $\alpha=0.0068 /{ }^{\circ} \mathrm{C}$
151975
Which statement is correct for the given circuit?
1 I through $R_{1}>$ I through $R_{2}$
2 I through $R_{3}>I$ through $R_{2}$ and $R_{1}$
3 I through $R_{2}>I$ through $R_{3}$ and $R_{1}$
4 I is same in $R_{1}, R_{2}$ and $R_{3}$
Explanation:
D In series current will always same. So, I will be same in $\mathrm{R}_{1}, \mathrm{R}_{2}, \mathrm{R}_{3}$.
VITEEE-2016
Current Electricity
151976
Dimensions of a block are $1 \mathrm{~cm} \times 1 \mathrm{~cm} \times 100 \mathrm{~cm}$. If specific resistance of its material is $3 \times 10^{-7} \Omega \mathrm{m}$, then the resistance between the opposite rectangular faces is
151977
Equal amounts of a metal are converted into cylindrical wires of different lengths $(L)$ and cross- sectional area (A). The wire with the maximum resistance is the one, which has
1 length $=\mathrm{L}$ and area $=\mathrm{A}$
2 length $=\frac{\mathrm{L}}{2}$ and area $=2 \mathrm{~A}$
3 length $=2 \mathrm{~L}$ and area $=\frac{\mathrm{A}}{2}$
4 all have the same resistance, as the amount of the metal is the same
Explanation:
C Resistance, $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ Where, $\rho$ is the resistivity, $\mathrm{R} \propto \frac{l}{\mathrm{~A}}$ According to question resistance $(\mathrm{R})$ will be maximum when- $l=2 \mathrm{~L}$ $\& \mathrm{~A}=\mathrm{A} / 2$
VITEEE-2008
Current Electricity
151978
The resistance of a field coil measures $50 \Omega$ at $20^{\circ} \mathrm{C}$ and $65 \Omega$ at $70^{\circ} \mathrm{C}$. The temperature coefficient of resistance is
1 $0.0086 /{ }^{\circ} \mathrm{C}$
2 $0.0068 /{ }^{\circ} \mathrm{C}$
3 $0.0096 /{ }^{\circ} \mathrm{C}$
4 $0.0999 /{ }^{\circ} \mathrm{C}$
Explanation:
B Given, Resistance at $20^{\circ} \mathrm{C}=50 \Omega$ Resistance at $70^{\circ} \mathrm{C}=65 \Omega$ $\because \quad \mathrm{R}=\mathrm{R}_{\mathrm{o}}(1+\alpha \mathrm{t})$ $\therefore \quad 50=\mathrm{R}_{\mathrm{o}}(1+\alpha 20)$ $65=\mathrm{R}_{\mathrm{o}}(1+\alpha 70)$ Dividing equation (ii) and equation (i), we get- $\frac{65}{50}=\frac{R_{o}(1+\alpha 70)}{R_{o}(1+\alpha 20)}$ $65+1300 \alpha=50+3500 \alpha$ $65-50=3500 \alpha-1300 \alpha$ $15=2200 \alpha$ $\alpha=\frac{15}{2200}$ $\alpha=0.0068 /{ }^{\circ} \mathrm{C}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Current Electricity
151975
Which statement is correct for the given circuit?
1 I through $R_{1}>$ I through $R_{2}$
2 I through $R_{3}>I$ through $R_{2}$ and $R_{1}$
3 I through $R_{2}>I$ through $R_{3}$ and $R_{1}$
4 I is same in $R_{1}, R_{2}$ and $R_{3}$
Explanation:
D In series current will always same. So, I will be same in $\mathrm{R}_{1}, \mathrm{R}_{2}, \mathrm{R}_{3}$.
VITEEE-2016
Current Electricity
151976
Dimensions of a block are $1 \mathrm{~cm} \times 1 \mathrm{~cm} \times 100 \mathrm{~cm}$. If specific resistance of its material is $3 \times 10^{-7} \Omega \mathrm{m}$, then the resistance between the opposite rectangular faces is
151977
Equal amounts of a metal are converted into cylindrical wires of different lengths $(L)$ and cross- sectional area (A). The wire with the maximum resistance is the one, which has
1 length $=\mathrm{L}$ and area $=\mathrm{A}$
2 length $=\frac{\mathrm{L}}{2}$ and area $=2 \mathrm{~A}$
3 length $=2 \mathrm{~L}$ and area $=\frac{\mathrm{A}}{2}$
4 all have the same resistance, as the amount of the metal is the same
Explanation:
C Resistance, $\mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ Where, $\rho$ is the resistivity, $\mathrm{R} \propto \frac{l}{\mathrm{~A}}$ According to question resistance $(\mathrm{R})$ will be maximum when- $l=2 \mathrm{~L}$ $\& \mathrm{~A}=\mathrm{A} / 2$
VITEEE-2008
Current Electricity
151978
The resistance of a field coil measures $50 \Omega$ at $20^{\circ} \mathrm{C}$ and $65 \Omega$ at $70^{\circ} \mathrm{C}$. The temperature coefficient of resistance is
1 $0.0086 /{ }^{\circ} \mathrm{C}$
2 $0.0068 /{ }^{\circ} \mathrm{C}$
3 $0.0096 /{ }^{\circ} \mathrm{C}$
4 $0.0999 /{ }^{\circ} \mathrm{C}$
Explanation:
B Given, Resistance at $20^{\circ} \mathrm{C}=50 \Omega$ Resistance at $70^{\circ} \mathrm{C}=65 \Omega$ $\because \quad \mathrm{R}=\mathrm{R}_{\mathrm{o}}(1+\alpha \mathrm{t})$ $\therefore \quad 50=\mathrm{R}_{\mathrm{o}}(1+\alpha 20)$ $65=\mathrm{R}_{\mathrm{o}}(1+\alpha 70)$ Dividing equation (ii) and equation (i), we get- $\frac{65}{50}=\frac{R_{o}(1+\alpha 70)}{R_{o}(1+\alpha 20)}$ $65+1300 \alpha=50+3500 \alpha$ $65-50=3500 \alpha-1300 \alpha$ $15=2200 \alpha$ $\alpha=\frac{15}{2200}$ $\alpha=0.0068 /{ }^{\circ} \mathrm{C}$
