The mass per unit length of a uniform wire… - JEE Main 2021 Physics PYQ

JEE Main 2021 — Physics Waves & Oscillations

2021integermedium

The mass per unit length of a uniform wire is $0.135g{\mathrm{cm}}^{-1}$ . A transverse wave of the form $y=-0.21\mathrm{sin}(x+30t)$ is produced in it, where $x$ is in meter and $t$ is in second. Then, the expected value of tension in the wire is $x\times {10}^{-2}N$ . Value of $x$ is (Round-off to the nearest integer)

Official previous-year question

Held on 26 Feb 2021 · Verified 6 Jul 2026.

Did you get this right?

Solution

$\mu =0.135\mathrm{gm}{\mathrm{cm}}^{-1}=0.0135\mathrm{kg}{m}^{-1}$

$y=-0.21\mathrm{sin}(x+30t)$

( $x$ in meter & $t$ in sec)

$v=\frac{\omega }{k}=\frac{30}{1}=30m{s}^{-1}$

$v=\sqrt{\frac{T}{\mu }}\Rightarrow T={v}^{2}\mu ={(30)}^{2}(0.0135)$

$=12.15$

$=x\times {10}^{-2}N$

$\Rightarrow x=1215$

Your note

JEE Main Waves & Oscillations in other years

Waves & Oscillations 2026 Waves & Oscillations 2025 Waves & Oscillations 2024 Waves & Oscillations 2023 Waves & Oscillations 2022 Waves & Oscillations 2020 Waves & Oscillations 2019 Waves & Oscillations 2018 Waves & Oscillations 2017 Waves & Oscillations 2016 Waves & Oscillations 2015 Waves & Oscillations 2014

More JEE Main Waves & Oscillations questions

The time period of a simple harmonic oscillator is $T=2 \pi \sqrt{\frac{k}{m}}$. Measured value of mass $(m)$ of the object is 10 g with an accuracy of 10 mg and time for 50 oscillations of the spring is found to be 60 s using a watch of 2 s resolution. Percentage error in determination of spring constant $(k)$ is $\_\_\_\_$ \%.

2026medium

Two waves of same frequency and amplitude travel in opposite directions. The resulting pattern is:

2026easy

A spring-mass system oscillates with angular frequency ω. If the mass is doubled and the spring constant is halved, the new angular frequency is:

2026hard

In an open organ pipe $\nu_{3}$ and $\nu_{6}$ are $3^{\text {rd }}$ and $6^{\text {th }}$ harmonic frequencies, respectively. If $\nu_{6}-\nu_{3}=2200 \mathrm{~Hz}$ then length of the pipe is $\_\_\_\_$ mm . (Take velocity of sound in air is $330 \mathrm{~m} / \mathrm{s}$.)

2026easy

A simple pendulum of string length 30 cm performs 20 oscillations in 10 s. The length of the string required for the pendulum to perform 40 oscillations in the same time duration is $\_\_\_\_$ cm. [Assume that the mass of the pendulum remains same.]

2026easy

View all Waves & Oscillations questions