1

### JEE Main 2019 (Online) 10th January Morning Slot

A string of length 1 m and mass 5 g is fixed at both ends. The tension in the string is 8.0 N. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to -
A
16.6 cm
B
10.0 cm
C
20.0 cm
D
33.3 cm

## Explanation

Velocity of wave on string

$V = \sqrt {{T \over \mu }} = \sqrt {{8 \over 5} \times 1000} = 40m/s$

Now, wavelength of wave

$\lambda = {v \over n} = {{40} \over {100}}m$

Separation b/w successive nodes,

${\lambda \over 2} = {{20} \over {100}}\,m$ $=$ 20 cm
2

### 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
3

### 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
4

### 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.