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JEE Advanced 2019 Paper 1

Section: Physics

Question : 7 of 54

Marks: +1, -0

If the ideal cell is replaced by a cell having internal resistance of 5Ω then the measured value of R will be more than 1000Ω
The measured value of R will be 978Ω < R < 982Ω
The resistance of the Voltmeter will be 100kΩ
The resistance of the Ammeter will be 0.02Ω(round off to 2nd decimal place)

Solution:

Option 4 :

(I_g)G=(I)(R_A)

(2 \times 10^{-6})10 = (1 \times 10^{-3})R_{A}

∴Resistance of ammeter

R_A=0.02 \Omega

(4) is correct

Option 3:

V=(I_g) \cdot R_V

\therefore R_{V}= \; \frac{V}{Ig}= \; \frac{10^{-1}}{2 \times 10^{-6}}=5 \times 10^{4} \Omega

=50 \text{k}\Omega

∴ (3) is wrong

Option 2:

R_{1}= \; \frac{R \cdot R_{V}}{R+R_{V}}, R_{2}=R_{A}

Reading of ammeter = I

Reading of voltmeter V = IR

Measured value of resistance is

R_{m}= \; \frac{V}{I}=R_{1}

= \; \frac{R \cdot R_V}{R+R_V}= \; \frac{10^{3} \times 5 \times 10^{4}}{10^{3}+5 \times 10^{4}}

= \; \frac{5 \times 10^{4}}{M}

=980.4 \Omega

∴ (2) is correct

Option 1: measured value of R depends on R and

R_V

only, but independent of

R_A

, r values.

∴ Measured value of resistance will again be 980.4Ω only

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