NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
151832
Resistance of wire at $20{ }^{\circ} \mathrm{C}$ is $10 \Omega$. At what temperature, resistance becomes $30 \Omega$ (The temperature coefficient of resistance $\alpha=$ $\left.0.0125 /{ }^{\circ} \mathrm{C}\right)$
1 $220^{\circ} \mathrm{C}$
2 $100{ }^{\circ} \mathrm{C}$
3 $180^{\circ} \mathrm{C}$
4 $300{ }^{\circ} \mathrm{C}$
Explanation:
C Given, Resistance of wire $\left(\mathrm{R}_{\mathrm{o}}\right)=10 \Omega$ Temperature $(\mathrm{T})=20^{\circ} \mathrm{C}=273+20=293 \mathrm{~K}$ Coefficient of resistance $(\alpha)=0.0125 /{ }^{\circ} \mathrm{C}$ $\mathrm{R}=30 \Omega, \Delta \mathrm{T}=(\mathrm{T}-293) \mathrm{K}$ $\because \quad \mathrm{R}=\mathrm{R}_{\mathrm{o}}(1+\alpha \Delta \mathrm{T})$ $30=10[1+0.0125(\mathrm{~T}-293)]$ $3=1+0.0125(\mathrm{~T}-293) 2$ $\frac{2}{0.0125}=\mathrm{T}-293$ $\mathrm{~T}=453 \mathrm{~K}=180^{\circ} \mathrm{C}$
TS EAMCET 30.07.2022
Current Electricity
151834
If resistance of a conductor is $5 \Omega$ at $50^{\circ} \mathrm{C}$ and $7 \Omega$ at $100^{\circ} \mathrm{C}$ the mean temperature coefficient of resistance of the material is
151835
Match the following? | Column-I | Column-II | | :--- | :--- | | A. Conductance | i. Gray | | B. Magnetic \ltbr> Induction | ii. Lumen | | C. Absorbed dose | iii. Tesla | | D. Luminous flux | iv. Siemens |
151836
Two resistors $R_{1}=(4 \pm 0.8) \Omega$ and $R_{2}=(4 \pm 0.4)$ $\Omega$ are connected in parallel. The equivalent resistance of their parallel combination will be
1 $(4 \pm 0.4) \Omega$
2 $(2 \pm 0.4) \Omega$
3 $(2 \pm 0.3) \Omega$
4 $(4 \pm 0.3) \Omega$
Explanation:
C Given, $\mathrm{R}_{1}=(4 \pm 0.8) \Omega, \mathrm{R}_{2}=(4 \pm 0.4) \Omega$ We know that, in parallel combination of resistors, $\frac{1}{\mathrm{R}_{\text {eq }}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $\mathrm{R}_{\text {eq }}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{4 \times 4}{4+4}=2 \Omega$ and also, $\frac{\Delta \mathrm{R}_{\mathrm{eq}}}{\mathrm{R}_{\mathrm{eq}}^{2}} =\frac{\Delta \mathrm{R}_{1}}{\mathrm{R}_{1}^{2}}+\frac{\Delta \mathrm{R}_{2}}{\mathrm{R}_{2}^{2}}$ $\frac{\Delta \mathrm{R}_{\mathrm{eq}}}{(2)^{2}} =\frac{0.8}{(4)^{2}}+\frac{0.4}{(4)^{2}}$ $\Delta \mathrm{R}_{\mathrm{eq}} =0.3 \Omega$ Hence, net equivalent resistance, $=\mathrm{R}_{\mathrm{eq}} \pm \Delta \mathrm{R}_{\mathrm{eq}}$ $=(2 \pm 0.3) \Omega$
151832
Resistance of wire at $20{ }^{\circ} \mathrm{C}$ is $10 \Omega$. At what temperature, resistance becomes $30 \Omega$ (The temperature coefficient of resistance $\alpha=$ $\left.0.0125 /{ }^{\circ} \mathrm{C}\right)$
1 $220^{\circ} \mathrm{C}$
2 $100{ }^{\circ} \mathrm{C}$
3 $180^{\circ} \mathrm{C}$
4 $300{ }^{\circ} \mathrm{C}$
Explanation:
C Given, Resistance of wire $\left(\mathrm{R}_{\mathrm{o}}\right)=10 \Omega$ Temperature $(\mathrm{T})=20^{\circ} \mathrm{C}=273+20=293 \mathrm{~K}$ Coefficient of resistance $(\alpha)=0.0125 /{ }^{\circ} \mathrm{C}$ $\mathrm{R}=30 \Omega, \Delta \mathrm{T}=(\mathrm{T}-293) \mathrm{K}$ $\because \quad \mathrm{R}=\mathrm{R}_{\mathrm{o}}(1+\alpha \Delta \mathrm{T})$ $30=10[1+0.0125(\mathrm{~T}-293)]$ $3=1+0.0125(\mathrm{~T}-293) 2$ $\frac{2}{0.0125}=\mathrm{T}-293$ $\mathrm{~T}=453 \mathrm{~K}=180^{\circ} \mathrm{C}$
TS EAMCET 30.07.2022
Current Electricity
151834
If resistance of a conductor is $5 \Omega$ at $50^{\circ} \mathrm{C}$ and $7 \Omega$ at $100^{\circ} \mathrm{C}$ the mean temperature coefficient of resistance of the material is
151835
Match the following? | Column-I | Column-II | | :--- | :--- | | A. Conductance | i. Gray | | B. Magnetic \ltbr> Induction | ii. Lumen | | C. Absorbed dose | iii. Tesla | | D. Luminous flux | iv. Siemens |
151836
Two resistors $R_{1}=(4 \pm 0.8) \Omega$ and $R_{2}=(4 \pm 0.4)$ $\Omega$ are connected in parallel. The equivalent resistance of their parallel combination will be
1 $(4 \pm 0.4) \Omega$
2 $(2 \pm 0.4) \Omega$
3 $(2 \pm 0.3) \Omega$
4 $(4 \pm 0.3) \Omega$
Explanation:
C Given, $\mathrm{R}_{1}=(4 \pm 0.8) \Omega, \mathrm{R}_{2}=(4 \pm 0.4) \Omega$ We know that, in parallel combination of resistors, $\frac{1}{\mathrm{R}_{\text {eq }}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $\mathrm{R}_{\text {eq }}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{4 \times 4}{4+4}=2 \Omega$ and also, $\frac{\Delta \mathrm{R}_{\mathrm{eq}}}{\mathrm{R}_{\mathrm{eq}}^{2}} =\frac{\Delta \mathrm{R}_{1}}{\mathrm{R}_{1}^{2}}+\frac{\Delta \mathrm{R}_{2}}{\mathrm{R}_{2}^{2}}$ $\frac{\Delta \mathrm{R}_{\mathrm{eq}}}{(2)^{2}} =\frac{0.8}{(4)^{2}}+\frac{0.4}{(4)^{2}}$ $\Delta \mathrm{R}_{\mathrm{eq}} =0.3 \Omega$ Hence, net equivalent resistance, $=\mathrm{R}_{\mathrm{eq}} \pm \Delta \mathrm{R}_{\mathrm{eq}}$ $=(2 \pm 0.3) \Omega$
151832
Resistance of wire at $20{ }^{\circ} \mathrm{C}$ is $10 \Omega$. At what temperature, resistance becomes $30 \Omega$ (The temperature coefficient of resistance $\alpha=$ $\left.0.0125 /{ }^{\circ} \mathrm{C}\right)$
1 $220^{\circ} \mathrm{C}$
2 $100{ }^{\circ} \mathrm{C}$
3 $180^{\circ} \mathrm{C}$
4 $300{ }^{\circ} \mathrm{C}$
Explanation:
C Given, Resistance of wire $\left(\mathrm{R}_{\mathrm{o}}\right)=10 \Omega$ Temperature $(\mathrm{T})=20^{\circ} \mathrm{C}=273+20=293 \mathrm{~K}$ Coefficient of resistance $(\alpha)=0.0125 /{ }^{\circ} \mathrm{C}$ $\mathrm{R}=30 \Omega, \Delta \mathrm{T}=(\mathrm{T}-293) \mathrm{K}$ $\because \quad \mathrm{R}=\mathrm{R}_{\mathrm{o}}(1+\alpha \Delta \mathrm{T})$ $30=10[1+0.0125(\mathrm{~T}-293)]$ $3=1+0.0125(\mathrm{~T}-293) 2$ $\frac{2}{0.0125}=\mathrm{T}-293$ $\mathrm{~T}=453 \mathrm{~K}=180^{\circ} \mathrm{C}$
TS EAMCET 30.07.2022
Current Electricity
151834
If resistance of a conductor is $5 \Omega$ at $50^{\circ} \mathrm{C}$ and $7 \Omega$ at $100^{\circ} \mathrm{C}$ the mean temperature coefficient of resistance of the material is
151835
Match the following? | Column-I | Column-II | | :--- | :--- | | A. Conductance | i. Gray | | B. Magnetic \ltbr> Induction | ii. Lumen | | C. Absorbed dose | iii. Tesla | | D. Luminous flux | iv. Siemens |
151836
Two resistors $R_{1}=(4 \pm 0.8) \Omega$ and $R_{2}=(4 \pm 0.4)$ $\Omega$ are connected in parallel. The equivalent resistance of their parallel combination will be
1 $(4 \pm 0.4) \Omega$
2 $(2 \pm 0.4) \Omega$
3 $(2 \pm 0.3) \Omega$
4 $(4 \pm 0.3) \Omega$
Explanation:
C Given, $\mathrm{R}_{1}=(4 \pm 0.8) \Omega, \mathrm{R}_{2}=(4 \pm 0.4) \Omega$ We know that, in parallel combination of resistors, $\frac{1}{\mathrm{R}_{\text {eq }}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $\mathrm{R}_{\text {eq }}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{4 \times 4}{4+4}=2 \Omega$ and also, $\frac{\Delta \mathrm{R}_{\mathrm{eq}}}{\mathrm{R}_{\mathrm{eq}}^{2}} =\frac{\Delta \mathrm{R}_{1}}{\mathrm{R}_{1}^{2}}+\frac{\Delta \mathrm{R}_{2}}{\mathrm{R}_{2}^{2}}$ $\frac{\Delta \mathrm{R}_{\mathrm{eq}}}{(2)^{2}} =\frac{0.8}{(4)^{2}}+\frac{0.4}{(4)^{2}}$ $\Delta \mathrm{R}_{\mathrm{eq}} =0.3 \Omega$ Hence, net equivalent resistance, $=\mathrm{R}_{\mathrm{eq}} \pm \Delta \mathrm{R}_{\mathrm{eq}}$ $=(2 \pm 0.3) \Omega$
151832
Resistance of wire at $20{ }^{\circ} \mathrm{C}$ is $10 \Omega$. At what temperature, resistance becomes $30 \Omega$ (The temperature coefficient of resistance $\alpha=$ $\left.0.0125 /{ }^{\circ} \mathrm{C}\right)$
1 $220^{\circ} \mathrm{C}$
2 $100{ }^{\circ} \mathrm{C}$
3 $180^{\circ} \mathrm{C}$
4 $300{ }^{\circ} \mathrm{C}$
Explanation:
C Given, Resistance of wire $\left(\mathrm{R}_{\mathrm{o}}\right)=10 \Omega$ Temperature $(\mathrm{T})=20^{\circ} \mathrm{C}=273+20=293 \mathrm{~K}$ Coefficient of resistance $(\alpha)=0.0125 /{ }^{\circ} \mathrm{C}$ $\mathrm{R}=30 \Omega, \Delta \mathrm{T}=(\mathrm{T}-293) \mathrm{K}$ $\because \quad \mathrm{R}=\mathrm{R}_{\mathrm{o}}(1+\alpha \Delta \mathrm{T})$ $30=10[1+0.0125(\mathrm{~T}-293)]$ $3=1+0.0125(\mathrm{~T}-293) 2$ $\frac{2}{0.0125}=\mathrm{T}-293$ $\mathrm{~T}=453 \mathrm{~K}=180^{\circ} \mathrm{C}$
TS EAMCET 30.07.2022
Current Electricity
151834
If resistance of a conductor is $5 \Omega$ at $50^{\circ} \mathrm{C}$ and $7 \Omega$ at $100^{\circ} \mathrm{C}$ the mean temperature coefficient of resistance of the material is
151835
Match the following? | Column-I | Column-II | | :--- | :--- | | A. Conductance | i. Gray | | B. Magnetic \ltbr> Induction | ii. Lumen | | C. Absorbed dose | iii. Tesla | | D. Luminous flux | iv. Siemens |
151836
Two resistors $R_{1}=(4 \pm 0.8) \Omega$ and $R_{2}=(4 \pm 0.4)$ $\Omega$ are connected in parallel. The equivalent resistance of their parallel combination will be
1 $(4 \pm 0.4) \Omega$
2 $(2 \pm 0.4) \Omega$
3 $(2 \pm 0.3) \Omega$
4 $(4 \pm 0.3) \Omega$
Explanation:
C Given, $\mathrm{R}_{1}=(4 \pm 0.8) \Omega, \mathrm{R}_{2}=(4 \pm 0.4) \Omega$ We know that, in parallel combination of resistors, $\frac{1}{\mathrm{R}_{\text {eq }}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $\mathrm{R}_{\text {eq }}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{4 \times 4}{4+4}=2 \Omega$ and also, $\frac{\Delta \mathrm{R}_{\mathrm{eq}}}{\mathrm{R}_{\mathrm{eq}}^{2}} =\frac{\Delta \mathrm{R}_{1}}{\mathrm{R}_{1}^{2}}+\frac{\Delta \mathrm{R}_{2}}{\mathrm{R}_{2}^{2}}$ $\frac{\Delta \mathrm{R}_{\mathrm{eq}}}{(2)^{2}} =\frac{0.8}{(4)^{2}}+\frac{0.4}{(4)^{2}}$ $\Delta \mathrm{R}_{\mathrm{eq}} =0.3 \Omega$ Hence, net equivalent resistance, $=\mathrm{R}_{\mathrm{eq}} \pm \Delta \mathrm{R}_{\mathrm{eq}}$ $=(2 \pm 0.3) \Omega$
