1

### 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

### 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

### 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

### 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

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

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