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A massless string connects two pulley of masses ' $2 \mathrm{~kg}$ ' and ' $1 \mathrm{~kg}$ ' respectively as shown in the figure.

The heavier pulley is fixed and free to rotate about its central axis while the other is free to rotate as well as translate. Find the acceleration of the lower pulley if the system was released from the rest. [Given, $g=10 \mathrm{~m} / \mathrm{s}^2$ ]
Option: 1
$\frac{4}{3} \mathrm{~gm} / \mathrm{s}^2$

Option: 2
$\frac{3}{2} \mathrm{~gm} / \mathrm{s}^2$

Option: 3
$\frac{3}{4} \mathrm{~gm} / \mathrm{s}^2$

Option: 4
$\frac{2}{3} \mathrm{~gm} / \mathrm{s}^2$

Not understanding sir

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Posted by

Raju vittal nandi

Calculate the acceleration of block $m_1$ of the following diagram. Assume all surfaces are frictionless . Here m₁ = 100kg and m₂ = 50kg

Option: 1
0.33m/s²

Option: 2
0.66m/s²

Option: 3
1m/s²

Option: 4
1.32m/s²

1.32

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Posted by

balda gayathri

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A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals µ. The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction µ and the distance x(=QR), are, respectively close to :
Option: 1 0.2 and 6.5 m
Option: 3 0.2 and 3.5 m
Option: 4 0.29 and 6.5 m

Work done by friction at QR = μmgx

In triangle, sin 30° = 1/2 = 2/PQ

PQ = 4 m

Work done by friction at PQ = μmg × cos 30° × 4 = μmg × √3/2 × 4 = 2√3μmg

Since work done by friction on parts PQ and QR are equal,

μmgx = 2√3μmg

x = 2√3 ≅ 3.5 m

Applying work energy theorem from P to R

decrease in P.E.=P.E.= loss of energy due to friction in PQPQ and QR

$\\ m g h=(\mu m g \cos \theta) P Q+\mu m g \times Q R\\ h=\mu \cos \theta \times P Q+\mu m g \times Q R\\ h=\mu \cos \theta \times P Q+\mu \times Q R =\mu \cos 30^{\circ} \times 4+\mu \times 2 \sqrt{3} =\mu\left(4 \times \frac{\sqrt{3}}{2}+2 \sqrt{3}\right)\\ h=\mu \times 4 \sqrt{3}\\ \mu=\frac{2}{4 \sqrt{3}}=\frac{1}{2 \sqrt{3}}=0.29$ where h=2(given)

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Posted by

Ritika Jonwal

A particle of mass m is moving along the side of a square of side ‘a’, with a uniform speed $\nu$ in the x-y plane as shown in the figure : Which of the following statements is false for the angular momentum $\vec{L}$ about the
origin?
Option: 1 $\vec{L}= -\frac{mv}{\sqrt{2}}R\hat{k}$ when the particle is moving from A to B.

Option: 2 $\vec{L}= mv\left [ \frac{R}{\sqrt{2}}-a \right ]\hat{k}$ when the particle is moving from C to D.

Option: 3 $\vec{L}= mv\left [ \frac{R}{\sqrt{2}}+a \right ]\hat{k}$ when the particle is moving from B to C.

Option: 4 $L=\frac{R}{\sqrt{2}} m v(-k)$ when the particle is moving from D to A.

$\\ In\ option\ (a), co-ordinates \ of\ A are \left(\frac{R}{\sqrt{2}}, \frac{R}{\sqrt{2}}\right) \\ \therefore \vec{r}=\left(\frac{R}{\sqrt{2}} \hat{i}+\frac{R}{\sqrt{2}} \hat{j}\right) and \vec{v}=v \hat{i}\\ \vec{L} m(\vec{r} \times \vec{v})=m\left(\frac{R}{\sqrt{2}} \hat{i}+\frac{R}{\sqrt{2}} \hat{j}\right) \times v \hat{i}\\ \vec{L}=-\frac{m R}{\sqrt{2}} v \hat{k}$

$\\ in\ option \ (b)\ it \ moves\ from \ C \ to\ D\\ L=\left(\frac{R}{\sqrt{2}}+a\right) m v(\hat{k})$

so option b is correct option

$\\ in \ option\ (c), \ For\ B$ \ to\ $C,$ \\ we have $L=\left(\frac{R}{\sqrt{2}}+a\right) m v(\hat{k})$

$\\ in \ option \ (d), \ When \ a \ particle\ is \ moving\ from \ D \ to\ A\\ L=\frac{R}{\sqrt{2}} m v(-k)$

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Posted by

Ritika Jonwal

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A body of mass m=10⁻² kg is moving in a medium and experiences a frictional force $F= -kv^{2}$ Its initial speed is $v_{0}= 10ms^{-1}$ . If, after 10 s, its energy is $\frac{1}{8}$ $mv{_{0}}^{2}$ the value of k will be :
Option: 1 10⁻³ kg m⁻¹
Option: 2 10⁻³ kg s⁻¹
Option: 3 10⁻⁴ kg m⁻¹
Option: 4 10⁻¹ kg m⁻¹ s⁻¹

As we learnt in

$\frac{1}{2}mv{_{f}}^{2}=\frac{1}{8}mv{_{o}}^{2}$

$v_{f}=\frac{Vo}{2}=5 m \slash s$

$\left ( 10^{-2} \right )\frac{dv}{dt}=-KV^{2}$

$\int_{10}^{5}=-100k\int_{0}^{10}dt\Rightarrow \frac{1}{5}-\frac{1}{10}=100 k (1^{\circ})$

$K=10^{-4}$

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Posted by

vishal kumar

Consider a uniform rod of mass $M=4m$ and length l pivoted about its centre. A mass m moving with velocity $v$ making angle $\theta =\frac{\pi }{4}$ to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod - mass system just after the collision is :
Option: 1 $\frac{4}{7}\frac{v}{l}$

Option: 2 $\frac{3\sqrt{2}}{7}\frac{v}{l}$

Option: 3 $\frac{3}{7\sqrt{2}}\frac{v}{l}$

Option: 4 $\frac{3}{7}\frac{v}{l}$

Let us conserve angular momentum about O:-

So, $L_i=\left ( \frac{mv}{\sqrt2} \right )\times \frac{l}{2}$ , where $\left ( \frac{mv}{\sqrt2} \right )$ is linear momentum and $\left ( \frac{l}{2} \right )$ is the distance from centre O.