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JEE Main 2021 (Online) 1st September Evening Shift

Numerical
A uniform heating wire of resistance 36$$\Omega$$ is connected across a potential difference of 240 V. The wire is then cut into half and potential difference of 240V is applied across each half separately. The ratio of power dissipation in first case to the total power dissipation in the second case would be 1 : x, where x is ____________
Your Input ________

Answer

Correct Answer is 4

Explanation

For Case I,

The potential difference of the uniform wire, V = 240 V

The resistance of the uniform wire, R1 = 36 $$\Omega$$

The power dissipation in the first case,

$${P_1} = {{{V^2}} \over {{R_1}}} = {{{{(240)}^2}} \over {36}}$$

For Case II,

The resistance of each half, $${R_2} = {{{R_1}} \over 2} = {{36} \over 2} = 18\Omega $$

$${P_2} = {{{V^2}} \over {{R_2}}} + {{{V^2}} \over {{R_2}}} = {{{{(240)}^2}} \over {18}} + {{{{(240)}^2}} \over {18}} = {{{{(240)}^2}} \over 9}$$

Thus, the ratio of the total power dissipation in the first case to the second case

$${{{P_1}} \over {{P_2}}} = {{{{(240)}^2}/36} \over {{{(240)}^2}/9}} \Rightarrow {{{P_1}} \over {{P_2}}} = {1 \over 4}$$

Comparing with, $${{{P_1}} \over {{P_2}}} = {1 \over x}$$

The value of the x = 4.
2

JEE Main 2021 (Online) 31st August Evening Shift

Numerical
A resistor dissipates 192 J of energy in 1s when a current of 4A is passed through it. Now, when the current is doubled, the amount of thermal energy dissipated in 5s in _________ J.
Your Input ________

Answer

Correct Answer is 3840

Explanation

E = i2Rt

192 = 16 (R) (1)

R = 12$$\Omega$$

E1 = (8)2 (12) (5)

= 3840 J
3

JEE Main 2021 (Online) 31st August Evening Shift

Numerical
A parallel plate capacitor of capacitance 200 $$\mu$$F is connected to a battery of 200 V. A dielectric slab of dielectric constant 2 is now inserted into the space between plates of capacitor while the battery remain connected. The change in the electrostatic energy in the capacitor will be ____________J.
Your Input ________

Answer

Correct Answer is 4

Explanation

$$\Delta U = {1 \over 2}(\Delta C){V^2}$$

$$\Delta U = {1 \over 2}(KC - C){V^2}$$

$$\Delta U = {1 \over 2}(2 - 1)C{V^2}$$

$$\Delta U = {1 \over 2} \times 200 \times {10^{ - 6}} \times 200 \times 200$$

$$\Delta U = 4$$ J
4

JEE Main 2021 (Online) 31st August Morning Shift

Numerical
A capacitor of 50 $$\mu$$F is connected in a circuit as shown in figure. The charge on the upper plate of the capacitor is ______________$$\mu$$C.

Your Input ________

Answer

Correct Answer is 100

Explanation



Potential Difference across each resistor = 2V

q = CV

= 50 $$\times$$ 10$$-$$6 $$\times$$ 2 = 100 $$\times$$ 10$$-$$6 = 100 $$\mu$$C

Questions Asked from Current Electricity

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