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2 months ago

Laws of Motion physics ncert exemplar solutions class 12th chapter two

In the arrangement shown in figure a₁, a₂, a₃ and a₄ are the accelerations of masses m₁, m₂, m₃ and m₄ respectively. Which of the following relation is true for this arrangement?

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Vishal Baghel

Contributor-Level 10

$\sum_{}^{} \vec{T} . \vec{a} = 0$

$\frac{T}{2} a_{1} + \frac{T}{4} a_{2} + \frac{T}{8} a_{3} + \frac{T}{8} a_{4} = 0$

$\Rightarrow 4 a_{1} + 2 a_{2} + a_{3} + a_{4} = 0$

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Laws of Motion Physics NCERT Exemplar Solutions Class 12th Chapter Five

A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2 rads^-1 in a horizontal plane about an axis vertical to its plane and passing through the centre of the ring. If two objects each of mass m be attached gently to the opposite ends of a diameter of ring, the will then rotate with an angular velocity (in rads^-1).

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Payal Gupta

Contributor-Level 10

By conservation of Angular momentum

L_i = L_f

$M R^{2} ω = (m R^{2} + 2 m R^{2}) ω'$

$\frac{2 M}{M + 2 m} = ω'$

$\Rightarrow$ $ω' = \frac{2 M}{M + 2 m}$

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Laws of Motion Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

A block of metal weighing 2kg is resting on a frictionless plane (as shown in figure). It is struck by a jet relasing water at a rate of 1 kgs^-1 and at a speed of 10 ms^-1. Then, the initial acceleration of the block, in ms^-2, will be:

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alok kumar singh

Contributor-Level 10

$a = \frac{F}{m} = \frac{m v}{m} = \frac{1 * 10}{2} = 5 m / s e c^{2}$

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Laws of Motion Physics NCERT Exemplar Solutions Class 12th Chapter Five

A body of mass M at rest explodes into pieces, in the ratio of masses 1 : 1: 2. Two smaller pieces fly off perpendicular to each other with velocities of 30 ms^-¹ and 40 ms-¹ respectively. The velocity of the third piece will be:

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alok kumar singh

Contributor-Level 10

Let 'v' be the velocity of third piece

com along x-axis

M * 0 = M₁v_x + M₃v'_x

$\Rightarrow 0 = M_{1} * 30 + M_{3} v' x$

$v_{x} 1 = \frac{- M_{1}}{M_{3}} * 30$

$\frac{- 1}{2} * 30$

$v_{x}^{1} = - 15 m / s e c$

Com along y-axis

&nb

...more

Let 'v' be the velocity of third piece

com along x-axis

M * 0 = M₁v_x + M₃v'_x

$\Rightarrow 0 = M_{1} * 30 + M_{3} v' x$

$v_{x} 1 = \frac{- M_{1}}{M_{3}} * 30$

$\frac{- 1}{2} * 30$

$v_{x}^{1} = - 15 m / s e c$

Com along y-axis

M * 0 = M₂v_y + M₃v'_y

v_y = $\frac{- M_{2}}{M_{3}} v_{y}$

v_y = $\frac{- 1}{2} * 40$

v_y = -20m/sec

^{$\vec{v} = - 15 \hat{i} - 20 \hat{j}$}

v = $\sqrt[]{1 5^{2} + 2 0^{2}} = \sqrt[]{225 + 400} = 25 m / s e c$

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Laws of Motion Physics NCERT Exemplar Solutions Class 12th Chapter Five

A body of mass M at rest explodes into pieces, in the ratio of masses 1 : 1: 2. Two smaller pieces fly off perpendicular to each other with velocities of 30 ms^-1 and 40 ms^-1 respectively. The velocity of the third piece will be

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Payal Gupta

Contributor-Level 10

Let 'v' be the velocity of third piece

com along x-axis

M * 0 = M₁v_x + M₃v'_x

$\Rightarrow 0 = M_{1} * 30 + M_{3} v' x$

$v_{x} 1 = \frac{- M_{1}}{M_{3}} * 30$

= $\frac{- 1}{2} * 30$

$v_{x}^{1} = - 15 m / s e c$

Com along y-axis

M * 0 = M₂vy + M₃v'y

v_y = $\frac{- M_{2}}{M_{3}} v_{y}$

v_y = $\frac{- 1}{2} * 40$

v_y = 20m/sec

$\vec{v} = - 15 \hat{i} - 20 \hat{j}$

v = $\sqrt[]{1 5^{2} + 2 0^{2}} = \sqrt[]{225 + 400} = 25 m / s e c$

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Laws of Motion Physics NCERT Exemplar Solutions Class 12th Chapter Five

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New answer posted

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Laws of Motion Physics NCERT Exemplar Solutions Class 12th Chapter Five

Two masses M₁ and M₂are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass M₂is twice that of M₁. The acceleration of the system is a₁. When the mass M₂ is thrice that of M₁ the acceleration of the system is a₂. The ratio $\frac{a_{1}}{a_{2}}$ will be:

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alok kumar singh

Contributor-Level 10

$a_{1} = \frac{m_{2} g - m_{1} g}{m_{1} + m_{2}} = \frac{2 m g - m g}{3 m} = \frac{g}{3}$

$a_{2} = \frac{3 m g - m g}{4 m} = \frac{g}{2}$

$\Rightarrow \frac{a_{1}}{a_{2}} = \frac{2}{3}$

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2 months ago

Laws of Motion Physics NCERT Exemplar Solutions Class 12th Chapter Five

A monkey of mass 50kg climbs on a rope which can withstand the tension (T) of 350 N. If monkey initially climbs down with an acceleration of 4 m/s² and then climbs up with an acceleration of 5 m/s². Choose the correct option (g = 10 m/s²).

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Vishal Baghel

Contributor-Level 10

For climbing downward

50 g – T = 50a

T = 500 – 50 * 4

= 300 N < 350 N

for climbing upwards

T – 50 g = 50 a

T – 500 = 50 * 5

T = 750 N > 350 N

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Laws of Motion Physics NCERT Exemplar Solutions Class 12th Chapter Five

Three masses M = 100kg, m₁ = 10kg and M₂ = 20kg are arranged in a system as shown in figure. All the surface are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force F is applied on the system so that the mass m₂ moves upward with an acceleration of 2ms^-2. The value of F is:

(Take g = 10ms^-2)

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Vishal Baghel

Contributor-Level 10

Given Acceleration of m₂ is 2m/s²

(upward)

N = 20a T – m₂g = m₂a

T – 20g = 20a

T = 200 + 20 * 2

T = 240 N

T = m₁ a₁

240 = 10a₁

a₁ = 24 m/s²

$\Rightarrow T a + T a_{1} - T a_{2} = 0$

$a_{2} = 24 + 2 = 26 m / s^{2}$

F – T – N = Ma₂

F – 240 = 100 * 26 + 20 * 26

F = 3360 N

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2 months ago

Laws of Motion Physics NCERT Exemplar Solutions Class 12th Chapter Five

A ball of mass 0.15 kg hits the wall with its initial speed of 12 ms^-1and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is 100 N, calculate the time during of the contact of ball with the wall.

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alok kumar singh

Contributor-Level 10

$F = \frac{Δ ρ}{Δ t} = \frac{1.8 - (- 1.8)}{Δ t} = 100$

$\Rightarrow Δ t = \frac{3.6}{100} = 0.036 s e c$

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