1

JEE Advanced 2016 Paper 2 Offline

MCQ (More than One Correct Answer)
In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a periodic motion is $$T = 2\pi \sqrt {{{7\left( {R - r} \right)} \over {5g}}} $$. The values of R and r are measured to be (60 $$ \pm $$ 1) mm and (10 $$ \pm $$ 1) mm, respectively. In five successive measurements, the time period is found to be 0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59 s. The least count of the watch used for the measurement of time period is 0.01 s. Which of the following statement(s) is(are) true?
A
The error in the measurement of r is 10 %
B
The error in the measurement of T is 3.57 %
C
The error in the measurement of T is 2 %
D
The error in the determined value of g is 11 %

Explanation

The time period is measured for five successive measurements as follows :

T(s) 0.52 0.56 0.57 0.54 0.59
$$\Delta $$T 0.04 0 0.01 0.02 0.03


Therefore,

$${T_{mean}} = {{0.52 + 0.56 + 0.57 + 0.54 + 0.59} \over 5}$$

$$ = {{2.78} \over 5} = 0.556 \approx 0.56$$

and we know $$\Delta$$T(absolute error) = $$\left| {{T_{mean}} - T} \right|$$

Error in reading = $$|{T_{mean}} - {T_1}| = 0.04$$
$$|{T_{mean}} - {T_2}| = 0.00$$
$$|{T_{mean}} - {T_3}| = 0.01$$
$$|{T_{mean}} - {T_4}| = 0.02$$
$$|{T_{mean}} - {T_5}| = 0.03$$

$$ \Rightarrow \Delta {T_{mean}} = {{0.04 + 0 + 0.01 + 0.02 + 0.03} \over 5} = 0.02$$

Thus, the error in the measurement of T is

$${{\Delta T} \over T} \times 100 = {{0.02} \over {0.56}} \times 100 = 3.57\% $$ .... (1)

Hence, option (B) is correct.

The error in the measurement of r is

$${{\Delta r} \over r} \times 100 = {1 \over {10}} \times 100 = 10\% $$ ..... (2)

Hence, option (A) is correct.

Now, it is given that

$$T = 2\pi \sqrt {{{7(R - r)} \over {5g}}} $$

$$ \Rightarrow {T^2} = 4{\pi ^2}\left( {{{7(R - r)} \over {5g}}} \right)$$

$$ \Rightarrow g = {{28{\pi ^2}} \over 5}\left( {{{R - r} \over {{T^2}}}} \right)$$

Taking log and differentiating on both the sides of this equation, we get

$$\ln g = \ln \left( {{{28{\pi ^2}} \over 5}} \right) + \ln (R - r) - 2\ln T$$

$$ \Rightarrow {{\Delta g} \over g} = \left( {{{\Delta R + \Delta r} \over {R - r}}} \right) + {{2\Delta T} \over T}$$

$$ = \left( {{{1 + 1} \over {50}}} \right) + 2 \times 0.0357$$

$$ \Rightarrow {{\Delta g} \over g} \times 100 = \left( {{1 \over {25}} + 2 \times 0.0357} \right) \times 100$$

$$ = (4 + 7.14)\% = 11.14\% = 11\% $$

Therefore, the error in the determined value g is 11%.

Answer (A), (B), (D)
2

JEE Advanced 2016 Paper 1 Offline

MCQ (More than One Correct Answer)
A length-scale (l) depends on the permittivity ($$\varepsilon $$) of a dielectric material, Boltzmann constant (kB), the absolute temperature (T), the number per unit volume (n) of certain charged particles, and the charge (q) carried by each of the particles. Which of the following expression(s) for I is(are) dimensionally correct?
A
$$l = \sqrt {\left( {{{n{q^2}} \over {\varepsilon {k_b}T}}} \right)} $$
B
$$l = \sqrt {\left( {{{\varepsilon {k_b}T} \over {n{q^2}}}} \right)} $$
C
$$l = \sqrt {\left( {{{{q^2}} \over {\varepsilon {n^{2/3}}{k_B}T}}} \right)} $$
D
$$l = \sqrt {\left( {{{{q^2}} \over {\varepsilon {n^{1/3}}{k_B}T}}} \right)} $$

Explanation

$$[n] = [{L^{ - 3}}];[q] = [AT]$$

$$[\varepsilon ] = [{M^{ - 1}}{L^{ - 3}}{A^2}{T^4}]$$

$$[T] = [L]$$

$$[l] = [L]$$

$$[{k_B}] = [{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}]$$

(a) RHS

$$ = \sqrt {{{[{L^{ - 3}}{A^2}{T^2}]} \over {[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]}}} $$

$$ = \sqrt {{{[{L^{ - 3}}{A^2}{T^2}]} \over {[{L^{ - 1}}{A^2}{T^2}]}}} = \sqrt {[{L^{ - 2}}]} = [{L^{ - 1}}]$$ Wrong

(b) RHS

$$ = \sqrt {{{[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]} \over {[{L^{ - 3}}{A^2}{T^2}]}}} $$

$$ = \sqrt {{{[{L^{ - 1}}{A^2}{T^2}]} \over {[{L^{ - 3}}{A^2}{T^2}]}}} = L$$ Correct

(c) RHS

$$ = \sqrt {{{[{A^2}{T^2}]} \over {[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{L^{ - 2}}][{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]}}} $$

$$ = \sqrt {[{L^3}]} $$ Wrong

(d) RHS

$$ = \sqrt {{{[{A^2}{T^2}]} \over {[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{L^{ - 1}}][{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]}}} $$

$$ = \sqrt {{{[{A^2}{T^2}]} \over {[{L^{ - 2}}{T^2}{A^2}]}}} = [L]$$ Correct
3

JEE Advanced 2015 Paper 2 Offline

MCQ (More than One Correct Answer)
In terms of potential difference V, electric current I, permittivity $${\varepsilon _0}$$, permeability $${\mu _0}$$ and speed of light c, the dimensionally correct equation(s) is(are)
A
$${\mu _0}{I^2} = {\varepsilon _0}{V^2}$$
B
$${\mu _0}I = {\mu _0}V$$
C
$$I = {\varepsilon _0}cV$$
D
$${\mu _0}cI = {\varepsilon _0}V$$
4

JEE Advanced 2015 Paper 1 Offline

MCQ (More than One Correct Answer)
Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:
A
If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.
B
If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.
C
If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.
D
If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.

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