1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 11th January Morning Slot

Equation of travelling wave on a stretched string of linear density 5 g/m is y = 0.03 sin(450 t – 9x) where distance and time are measured in SI units. The tension in the string is :
A
10 N
B
7.5 N
C
5 N
D
12.5 N

Explanation

y = 0.03 sin(450 t $$-$$ 9x)

v = $${\omega \over k} = {{450} \over 9}$$ = 50m/s

v = $$\sqrt {{T \over \mu }} \Rightarrow {T \over \mu }$$ = 2500

$$ \Rightarrow $$  T = 2500 $$ \times $$ 5 $$ \times $$ 10$$-$$3

= 12.5 N
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 12th January Evening Slot

A resonance tube is old and has jagged end. It is still used in the laboratory to determine velocity of sound in air. A tuning fork of frequency 512 Hz produces first resonance when the tube is filled with water to a mark 11 cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256 Hz which produces first resonance when water reaches a mark 27 cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to :
A
335 ms–1
B
328 ms–1
C
341 ms–1
D
322 ms–1
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 8th April Morning Slot

A wire of length 2L, is made by joining two wires A and B of same length but different radii r and 2r and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire A is p and that in B is q then the ratio p : q is :
A
3 : 5
B
4 : 9
C
1 : 2
D
1 : 4

Explanation

Let mass per unit length of wires are $$\mu $$1 and $$\mu $$2 respectively

$$ \because $$ Materials are same, so density $$\rho $$ will be same.

$$ \therefore $$ $${\mu _1} = {{\rho \pi {r^2}L} \over L} = \mu $$ and $${\mu _2} = {{\rho 4\pi {r^2}L} \over L} = 4\mu $$

Tension in both are same = T, let speed of wave in wires are V1 and V2

$${V_1} = {{{V_1}} \over {2L}} = {V \over {2L}}\,\,\& \,{V_2} = {{{V_2}} \over {2L}} = {V \over {4L}}$$

Frequency at which both resonate is L.C.M. of both frequencies (i.e : $${V \over {2L}}$$ )

Hence number of loops in wires are 1 and 2 respectively

So, ratio of number of antinodes is 1 : 2.
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 8th April Evening Slot

A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to 1/1000 of the original amplitude is close to :-
A
100 s
B
10 s
C
20 s
D
50 s

Explanation

Time for 10 oscillations = $${{10} \over 5} = 2\,s$$

A = A0 e–kt

$${1 \over 2} = {e^{ - 2k}} \Rightarrow \ln 2 = 2k$$

10–3 = e–kt $$ \Rightarrow $$ 3In10 = kt

$$t = {{3\ln 10} \over k} = {{3\ln 10} \over {\ln 2}} \times 2$$

= $$6 \times {{2.3} \over {0.69}} \approx 20\,s$$

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