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### JEE Main 2017 (Online) 8th April Morning Slot

A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an 8 kg block placed on the same table. So, the frequency of vibration of the 8 kg block is :
A
${1 \over 4}Hz$
B
${1 \over {2\sqrt 2 }}Hz$
C
${1 \over 2}Hz$
D
$2$ $Hz$

## Explanation

For 1 kg block : Here frequency of spring (f) = ${1 \over {2\pi }}\sqrt {{k \over m}}$

Given that, F = 1 Hz

$\therefore\,\,\,$ ${1 \over {2\pi }}\sqrt {{k \over 1}}$ = 1

$\Rightarrow $$\,\,\, k = 4\pi 2 N m-1 For 8 kg Block : Here two identical springs are attached in parallel. So, Keq = k + k = 2k \therefore\,\,\, Frequency of 8 kg block, F' = {1 \over {2\pi }}\sqrt {{{{k_{eq}}} \over {m'}}} = {1 \over {2\pi }}\sqrt {{{2k} \over 8}} = {1 \over {2\pi }}\sqrt {{{2 \times 4{\pi ^2}} \over 8}} = {1 \over {2\pi }} \times \pi = {1 \over 2} Hz 2 MCQ (Single Correct Answer) ### JEE Main 2017 (Online) 8th April Morning Slot The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s−1 . At, t = 0 the displacement is 5 m. What is the maximum acceleration ? The initial phase is {\pi \over 4}. A 500 m/s2 B 500 \sqrt 2 m/ s2 C 750 m/s2 D 750 \sqrt 2 m / s2 ## Explanation Mximum velocity, Vmax = a\omega Maximum acceleration, Amax = a\omega 2 Given that, {{a{\omega ^2}} \over {a\omega }} = 10 \Rightarrow$$\,\,\,$ $\omega$ = 10 s$-$1

Displacement, x = a sin ($\omega$t + ${\pi \over 4}$)

at t = 0, displacement x = 5

$\therefore\,\,\,$ 5 = a sin $\left( {{\pi \over 4}} \right)$

$\Rightarrow $$\,\,\, 5 = a \times {1 \over {\sqrt 2 }} \Rightarrow$$\,\,\,$ a = 5${\sqrt 2 }$

$\therefore\,\,\,$ Maximum acceleration,

Amax = a$\omega$2 = 5 ${\sqrt 2 }$ $\times$ (10)2

= 500 ${\sqrt 2 }$ m/s2
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### JEE Main 2017 (Online) 9th April Morning Slot

A block of mass 0.1 kg is connected to an elastic spring of spring constant 640 Nm−1 and oscillates in a damping medium of damping constant 10−2 kg s−1 . The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to :
A
2 s
B
3.5 s
C
5 s
D
7 s
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### JEE Main 2018 (Online) 15th April Morning Slot

A thin uniform tube is bent into a circle of radius $r$ in the vertical plane. Equal volumes of two immiscible liquids, whose densities are ${\rho _1}$ and ${\rho _2}$ $\left( {{\rho _1} > {\rho _2}} \right),$ fill half the circle. The angle $\theta$ between the radius vector passing through the common interface and the vertical is :
A
$\theta = {\tan ^{ - 1}}\pi \left( {{{{\rho _1}} \over {{\rho _2}}}} \right)$
B
$\theta = {\tan ^{ - 1}}{\pi \over 2}\left( {{{{\rho _1}} \over {{\rho _2}}}} \right)$
C
$\theta = {\tan ^{ - 1}}\left( {{{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}} \right)$
D
$\theta = {\tan ^{ - 1}}{\pi \over 2}\left( {{{{\rho _1} + {\rho _2}} \over {{\rho _1} - {\rho _2}}}} \right)$

## Explanation As system is in equilibrium so the pressuse at A from both side of the liquid must be equal .

(r cos $\theta$ + r sin $\theta$) $\rho$2g = (r cos $\theta$ $-$ r sin $\theta$) $\rho$1g

$\Rightarrow $$\,\,\, {{{\rho _1}} \over {{\rho _2}}} = {{\sin \theta + \cos \theta } \over {\cos \theta - \sin \theta }} = {{1 + \tan \theta } \over {1 - \tan \theta }} \Rightarrow$$\,\,\,$ $\rho$1 $-$ $\rho$1 tan$\theta$ = $\rho$2 + $\rho$2 tan$\theta$

$\Rightarrow $$\,\,\, (\rho 1 + \rho 2) tan\theta = \rho 1 - \rho 2 \Rightarrow$$\,\,\,$ tan$\theta$ = ${{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}$

$\Rightarrow$$\,\,\,$ $\theta$ = tan$-$1 $\left( {{{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}} \right)$