1
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 16th April Morning Slot

The end correction of a resonance column is 1 cm. If the shortest length resonating with the tunning fork is 10 cm, the next resonating length should be :
A
28 cm
B
32 cm
C
36 cm
D
40 c

Explanation

Given, End correction (e) = 1 cm

For first resonance,

$${\lambda \over 4} = {l_1} + e$$ = 10 + 1 = 11 cm

For second resonance,

$${3\lambda \over 4} = {l_2} + e$$

$$ \Rightarrow $$ $${l_2}$$ = 3 $$ \times $$ 11 - 1 = 32 cm
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16. The intensity of the waves are in the ratio :
A
16 : 9
B
25 : 9
C
4 : 1
D
5 : 3

Explanation

Given that,

$${{{{\rm I}_{\max }}} \over {{{\mathop{\rm I}\nolimits} _{min}}}} = {{16} \over 1}$$

We know,

Imax $$=$$ $${\left( {\sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} } \right)^2}$$

and Imin $$ = {\left( {\sqrt {{{\rm I}_1}} - \sqrt {{{\rm I}_2}} } \right)^2}$$

$$ \therefore $$   $${{{{\left( {\sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} } \right)}^2}} \over {{{\left( {\sqrt {{{\rm I}_1}} - \sqrt {{{\rm I}_2}} } \right)}^2}}} = {{16} \over 1}$$

$$ \Rightarrow $$   $${{\sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} } \over {\sqrt {{{\rm I}_1}} - \sqrt {{{\rm I}_2}} }} = {4 \over 1}$$

$$ \Rightarrow $$   $$4\sqrt {{{\rm I}_1}} - 4\sqrt {{{\rm I}_2}} = \sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} $$

$$ \Rightarrow $$   $$3\sqrt {{{\rm I}_1}} = 5\sqrt {{{\rm I}_2}} $$

$$ \Rightarrow $$   $${{\sqrt {{{\rm I}_1}} } \over {\sqrt {{{\rm I}_2}} }} = {5 \over 3}$$

$$ \Rightarrow $$   $${{{{\rm I}_1}} \over {{{\rm I}_2}}} = {{25} \over 9}$$
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of radio m/M is close to :
A
0.77
B
0.57
C
0.37
D
0.17

Explanation

Initially :



After putting 2 masses of each 'm' at a distance $${L \over 2}$$ from center :



We know,

Time period (T) = 2$$\pi $$ $$\sqrt {{{\rm I} \over C}} $$

$$ \therefore $$  T $$ \propto $$ $$\sqrt {\rm I} $$

$$ \therefore $$   Frequency (f) $$ \propto $$ $$\sqrt {{1 \over {\rm I}}} $$

$$ \therefore $$   $${{{f_1}} \over {{f_2}}}$$ = $$\sqrt {{{{{\rm I}_2}} \over {{{\rm I}_1}}}} $$

Also given that,
After putting two masses 'm' at both end new frequency becomes 80% of initial frequency.

$$ \therefore $$   f2 = 0.8f1

$$ \therefore $$   $${{{f_1}} \over {0.8{f_1}}}$$ = $$\sqrt {{{{{\rm I}_2}} \over {{{\rm I}_1}}}} $$

$$ \therefore $$   $${{{{{\rm I}_1}} \over {{{\rm I}_2}}}}$$ = 0.64

Initial moment of inertia of the system,

$${{{\rm I}_1}}$$ = $${{M{{\left( {2L} \right)}^2}} \over {12}}$$

Final moment of inertia of the system,

I2 = $${{M{{\left( {2L} \right)}^2}} \over {12}}$$ + 2$$\left( {m{{\left( {{L \over 2}} \right)}^2}} \right)$$

$$ \therefore $$   $${{M{{\left( {2L} \right)}^2}} \over {12}}$$ = 0.64 $$\left[ {{{M{L^2}} \over 3} + {{m{L^2}} \over 2}} \right]$$

$$ \Rightarrow $$   $${{M{L^2}} \over {3 \times 0.64}}$$ = $${{M{L^2}} \over 3}$$ + $${{M{L^2}} \over 2}$$

$$ \Rightarrow $$   $${M \over {1.92}} - {M \over 3} = {m \over 2}$$

$$ \Rightarrow $$   $${{1.08M} \over {3 \times 1.92}}$$ = $${m \over 2}$$

$$ \Rightarrow $$   $${m \over M}$$ = $${{1.08 \times 2} \over {3 \times 1.92}}$$ = 0.37
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

A musician using an open flute of length 50 cm producess second harmonic sound waves. A person runs towards the musician from another end of hall at a speed of 10 km/h. If the wave speed is 330 m/s, the frequency heard by the running person shall be close to :
A
666 Hz
B
753 Hz
C
500 Hz
D
333 Hz

Explanation

Frequency of sound wave produce by flute

= $${{2{V_S}} \over {2\ell }}$$

= $${{2 \times 330} \over {2 \times 50 \times {{10}^{ - 2}}}}$$

= 660 Hz

Speed of the observer

= 10km/hr

= 10 $$ \times $$ $${5 \over {18}}$$ m/s

= $${{25} \over 9}$$ m/s

Frequency heard by the observer,

f' = $$\left( {{{{V_S} + {V_o}} \over {{V_S}}}} \right)f$$

= $$\left( {{{330 + {{25} \over 9}} \over {330}}} \right) \times 660$$

= 666 Hz

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