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*Simulator*

**Previous Years AIEEE/JEE Main Questions**

**Statement–1 :**The temperature dependence of resistance is usually given as $\mathrm{R}=\mathrm{R}_{0}(1+\alpha \Delta \mathrm{t})$. The resistance of a wire changes from $100 \Omega$ to $150 \Omega$ when its temperature is increased from $27^{\circ} \mathrm{C}$ to $227^{\circ} \mathrm{C}$. This implies that = $2.5 \times 10^{-3} /^{\circ} \mathrm{C}$.

**Statement–2 : **$\mathrm{R}=\mathrm{R}_{0}(1+\alpha \Delta \mathrm{t})$ is valid only when the change in the temperature $\Delta \mathrm{T}$ is small and $\Delta \mathrm{R}=\left(\mathrm{R}-\mathrm{R}_{0}\right)<<\mathrm{R}_{0}$.

(1) Statement–1 is true, Statement–2 is true; Statement–2 is not the correct explanation of Statement–1

(2) Statement–1 is false, Statement–2 is true

(3) Statement–1 is true, Statement–2 is false

(4) Statement–1 is true, Statement–2 is true; Statement–2 is the correct explanation of Statement–1

**[AIEEE – 2009]**

**Sol.**(1)

(1) $\frac{\alpha_{1}+\alpha_{2}}{2}, \frac{\alpha_{1}+\alpha_{2}}{2}$

(2) $\frac{\alpha_{1}+\alpha_{2}}{2}, \alpha_{1}+\alpha_{2}$

(3) $\alpha_{1}+\alpha_{2}, \frac{\alpha_{1}+\alpha_{2}}{2}$

(4) $\alpha_{1}+\alpha_{2}, \frac{\alpha_{1} \alpha_{2}}{\alpha_{1}+\alpha_{2}}$

**[AIEEE – 2010]**

**Sol.**(1)

For series combination

$\alpha_{\mathrm{S}}=\frac{\alpha_{1} \mathrm{R}_{01}+\alpha_{2} \mathrm{R}_{02}}{\mathrm{R}_{01}+\mathrm{R}_{02}}$

$\mathrm{R}_{01}=\mathrm{R}_{02}=\mathrm{R}_{0}(\text { given })$

$\alpha_{\mathrm{S}}=\frac{\alpha_{1}+\alpha_{2}}{2}$

For parallel combination

$\frac{1}{\mathrm{R}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$

$\Rightarrow \frac{1}{\mathrm{R}_{e q}}=\frac{1}{\mathrm{R}_{0}\left(1+\alpha_{1} \mathrm{t}\right)}+\frac{1}{\mathrm{R}_{0}\left(1+\alpha_{2} \mathrm{t}\right)}$

$\frac{1}{\frac{\mathrm{R}_{0}}{2}\left(1+\alpha_{p} \mathrm{t}\right)}=\frac{1}{\mathrm{R}_{0}\left(1+\alpha_{1} \mathrm{t}\right)}+\frac{1}{\mathrm{R}_{0}\left(1+\alpha_{2} \mathrm{t}\right)}$

$2\left(1+\alpha_{p} t\right)^{-1}=\left(1+\alpha_{1} t\right)^{-1}+\left(1+\alpha_{2} t\right)^{-1}$

using binomial expansion

$2-2 \alpha_{\mathrm{p}} \mathrm{t}=1-\alpha_{1} \mathrm{t}+1-\alpha_{2} \mathrm{t} \Rightarrow \alpha_{\mathrm{p}}=\frac{\alpha_{1}+\alpha_{2}}{2}$

(1) decrease by 0.2%

(2) decrease by 0.05%

(3) increase by 0.05%

(4) increase by 0.2%

**[AIEEE – 2011]**

**Sol.**(4)

$\mathrm{R}=\rho \frac{\ell}{\mathrm{A}} \Rightarrow \mathrm{R} \alpha \ell^{2}$

$\frac{\Delta \mathrm{R}}{\mathrm{R}}=\frac{\Delta \mathrm{R}}{\mathrm{R}}=\frac{2 \Delta \ell}{\ell}=2[0.1]=0.2 \%$ increase.

(1) 20% (2) 5% (3) 10% (4) 15%

**[AIEEE – 2011]**

**Sol.**(2)

(1) 0.2 V/m (2) 1 V/m (3) 0.5 V/m (4) 0.1 V/m

**[AIEEE – 2011]**

**Sol.**(4)

(1) 3% (2) 6% (3) zero (4) 1%

**[AIEEE – 2012]**

**Sol.**(2)

(1) Neither (2) Both (3) 100 W (4) 25 W

**[AIEEE – 2012]**

**Sol.**(4)

(1) zero Volt (2) 2.9 Volt (3) 13.3 Volt (4) 10.04 Volt

**[JEE-Mains 2013]**

**Sol.**(4)

**Statement-I :** Higher the range, greater is the resistance of ammeter.

**Statement-II : **To increase the range of ammeter, additional shunt needs to be used across it.

(1) Statement-I is true, Statement-II is true, Statement-II is the **correct **explanation of Statement- I

(2) Statement-I is true, Statement-II is true, Statement-II is **not **the correct explanation of Statement- I.

(3) Statement-I is **true,** Statement-II is false.

(4) Statement-I is **false,** Statement-II is true.

**[JEE-Mains 2013]**

**Sol.**(4)

To increase the range of ammeter, resistance should be decreased (So additional shunt connected in parallel) so total resistance to ammeter decreases.

(1) 12 A (2) 14 A (3) 8 A (4) 10 A

**[JEE-Mains 2014]**

**Sol.**(1)

All devices are in parallel so total current drawn is gives as

(1) $1.6 \times 10^{-6} \Omega \mathrm{m}$

(2) $1.6 \times 10^{-5} \Omega \mathrm{m}$

(3) $1.6 \times 10^{-8} \Omega \mathrm{m}$

(4) $1.6 \times 10^{-7} \Omega \mathrm{m}$

**[JEE-Mains 2015]**

**Sol.**(2)

$\mathrm{V} \quad=\mathrm{i} \mathrm{R}=\left(\mathrm{neAV}_{\mathrm{d}}\right)\left(\frac{\rho \ell}{\mathrm{A}}\right)$

$\mathrm{V}=\mathrm{neV}_{\mathrm{d}} \ell$

$\rho=\frac{\mathrm{V}}{\mathrm{neV}_{\mathrm{d}} \ell}$

On putting values are got the answer

$=1.6 \times 10^{-5} \Omega \mathrm{m}$

(1) 0.13 A, from Q to P

(2) 0.13 A, from P to Q

(3) 1.3 A, from P to Q

(4) 0A

**[JEE-Mains 2015]**

**Sol.**(1)

(1) $3 \Omega$

(2) $0.01 \Omega$

(3) $2 \Omega$

(4) $0.1 \Omega$

**[JEE-Mains 2016]**

**Sol.**(2)

(1) A rheostat can be used as a potential divider

(2) Kirchhoff’s second law represents energy conservation

(3) Wheatstone bridge is the most sensitive when all the four resistances are of the same order

of magnitude.

(4) In a balanced wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed.

**[JEE-Mains 2017]**

**Sol.**(4)

On interchanging Cell & Galvanometer.

On balancing condition

$\frac{\mathrm{R}_{1}}{\mathrm{R}_{3}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{4}}$ ….(1)

(1) $2.535 \times 10^{3} \Omega$

(2) $4.005 \times 10^{3} \Omega$

(3) $1.985 \times 10^{3} \Omega$

(4) $2.045 \times 10^{3} \Omega$

**[JEE-Mains 2017]**

**Sol.**(3)

$10=\left(5 \times 10^{-3}\right)(15+\mathrm{R})$

$r=1985 \Omega$

(1) 0.5 A (2) 0 A (3) 1 A (4) 0.25 A

**[JEE-Mains 2017]**

**Sol.**(2)

Taking voltage of point A as = 0

Then voltage at other points can be written as shown in figure

Hence voltage across all resistance is zero.

Hence current = 0

(1) $1.5 \Omega$

(2) $2 \Omega$

(3) $2.5 \Omega$

(4) 1 $\Omega$

**[JEE-Mains 2018]**

**Sol.**(1)

without shunting condition :

On balancing

$\mathrm{E}_{\mathrm{s}}-\frac{\mathrm{E}_{\mathrm{s}}}{(\mathrm{r}+\mathrm{R})} \mathrm{r}=40 \times \mathrm{x} \quad \ldots .(2)$

On solving :

(1) $505 \mathrm{k\Omega}$

(2) $550 \mathrm{k} \Omega$

(3) $910 \mathrm{k} \Omega$

(4) $990 \mathrm{k} \Omega$

**[JEE-Mains 2018]**

**Sol.**(2)

$\mathrm{R}_{1}+\mathrm{R}_{2}=1000 \Rightarrow \mathrm{R}_{2}=1000-\mathrm{R}_{1}$

On balancing condition

$\mathrm{R}_{1}(100-\ell)=\left(1000-\mathrm{R}_{1}\right) \ell \quad \ldots(1)$

On Inter changing resistance

On balancing condition

(1) 11.5 V and 11.6 V

(2) 11.4 V and 11.5 V

(3) 11.7 V and 11.8 V

(4) 11.6 V and 11.7 V

** [JEE-Mains 2018]**

**Sol.**(1)

in the last question, the values for the internal resistances are wrong. different values are used in the solution.

I was very helpful to me🤗

Thank u

Excellent problems and for my side one request that is please keep more problems

its very nice tq sir

sir last q me error h

I wanted to notify that the Last question has a printing error. The 12V battery has an internal resistance of 1 ohm and not 10. Please rectify this error. Thank you

Current Electricity is Easy scoring topic

Sir last question me r 12v ka Galt print hai