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4 days ago

Laws of Motion Class 11th

A block rests on an inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between block and the plane is 0.8. If the frictional force on the block is 10 N, then mass of block (in kg) (Take g = 10 m/s² )

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alok kumar singh

Contributor-Level 10

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New answer posted

4 days ago

Laws of Motion Class 11th

Bullets of 0.03 kg mass each hit a plate at a rate of 200 bullets per second with velocity of 50 m/s and reflect back with velocity of 30 m/s. The average force acting on the plate in newton is

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alok kumar singh

Contributor-Level 10

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New answer posted

6 days ago

Laws of Motion Class 11th

The root mean square speed of molecules of a gas is 1260 m/s. The average speed of the gas molecules is

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alok kumar singh

Contributor-Level 10

= 0.92 * 1260 = 1161 m/s

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a week ago

Laws of Motion Class 11th

Find the acceleration of 2 kg block shown in the diagram. (neglect friction)

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alok kumar singh

Contributor-Level 10

For 2 kg block

T – 2g sin37 = 2a . (i)

For 4 kg block

4g – 2T = $\frac{4 a}{2}$

2g – T = a . (ii)

T = (2g – a)

2g – a – 2g * $\frac{3}{5}$ = 2a

3a = 2g * $\frac{2}{5}$

$a = \frac{4 g}{15}$

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a week ago

Laws of Motion Class 11th

In the system shown below, the pulley 4 string are ideal. If the acceleration of blocks is $\frac{g}{8}$ , find $\frac{m_{1}}{m_{2}}$

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alok kumar singh

Contributor-Level 10

$a = \frac{(m_{1} - m_{2}) g}{(m_{1} + m_{2})} = \frac{g}{8}$

8m₁ – 8m₂ = m₁ + m₂

7m₁ = 9m₂

$\frac{m_{1}}{m_{2}} = \frac{9}{7}$

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a week ago

Laws of Motion Class 11th

For the block shown, F₁ is the minimum force required to move block upwards and F₂ is the minimum force required to prevent it from slipping, find $| {\vec{F}}_{1} - {\vec{F}}_{2} |$

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alok kumar singh

Contributor-Level 10

f_K = μmg cosq

= 0.1 * $\frac{50 * \sqrt[]{3}}{2}$

$= 2.5 \sqrt[]{3} N$

F₁= mg sinq + f_K

= 25 + 2.5 $\sqrt[]{3}$

F₂= mg sinq – f_K

= 25 – 2.5

F₁ – F₂ = 5 $\sqrt[]{3}$ N

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2 weeks ago

Laws of Motion Class 11th

For a given system find out $T_{1}'$ and $T_{2}'$ (forces exerted by the pulley on the support) respectively.

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alok kumar singh

Contributor-Level 10

From F.B. Ds of

$9 k g \to 90 - T_{1} = 9 a \dots . (1)$

$6 k g \to T_{1} - T_{2} = 6 a \dots . . (2)$

$5 k g \to T_{2} - 50 = 5 a \dots . . (3)$

$(1) + (2) + (3)$

$40 = 20 a ? a = 2 m / s^{2}$

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2 weeks ago

Laws of Motion Class 11th

A ball is projected from ground with initial momentum p at an angle θ . What is the magnitude of net change in linear momentum when it reaches at maximum height of journey?

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alok kumar singh

Contributor-Level 10

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New answer posted

2 weeks ago

Laws of Motion Class 11th

When an elevator is moving downward with a increasing speed, the apparent weight of a body inside elevator

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alok kumar singh

Contributor-Level 10

Apparent weight = mg – ma

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2 weeks ago

Laws of Motion Class 11th

Two masses A and B, each of mass M are fixed together by a massless spring. A force acts on the mass B as shown in figure. If the mass A starts moving away from mass B with acceleration 'a', then the acceleration of mass B will be:

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alok kumar singh

Contributor-Level 10

Let the spring is in extension state and $a_{B} > a_{A}, {\vec{a}}_{A B} = - a_{A} \hat{i} - (- a_{B} \hat{i}) = (- a_{A} + a_{B}) \hat{i}$

Hence we can say that block moves away from block B in the frame of B

F – kx = Ma_B . (1)

kx = Ma . (2)

$\Rightarrow a_{B} = \frac{F}{M} - a$

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