Question

Two masses m₁ = 5 kg and m₂ = 10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of the horizontal surface is 0.15. The minimum weight m that should be put on top of m₂ to stop the motion is :

10.3 kg

18.3 kg

27.3 kg

43.3 kg

Solution

C.

27.3 kg

Given: m₁ = 5kg;
m₂ = 10 kg
μ = 0.15
For m₁, m₁g -T =m₁a
= 50-T = 5 x a
and T - 0.15 (m+10) g = (10 + m)a
For rest a = 0
or 50 = 0.15 (m +10) 10

$\Rightarrow 5 = \frac{3}{20} (m + 10) \frac{100}{3} = m + 10 m = 23.3 k g$

The minimum value from the options, satisfying the above condition is, m = 27.3 kg

Question

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In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after the collision, is:

$\frac{ν_{o}}{\sqrt{2}}$

v_o/4

$\sqrt{2} v_{0}$

v₀/2

Solution

C.

$\sqrt{2} v_{0}$

$\frac{1}{2} m v_{1}^{2} + \frac{1}{2} m v_{2}^{2} = \frac{3}{2} (\frac{1}{2} m v_{0}^{2}) \Rightarrow v_{1}^{2} + v_{2}^{2} = \frac{3}{2} v_{0}^{2} . . . . (i) F r o m m o m e n t u m c o n s e r v a t i o n m v_{0} = m (v_{1} + v_{2}) . . . (i i) S q u a r r i n g b o t h s i d e s, (v_{1} + v_{2})^{2} = v_{0}^{2} \Rightarrow v_{1}^{2} + v_{2}^{2} + 2 v_{1} v_{2} = v_{0}^{2} 2 v_{1} v_{2} = - \frac{v_{0}^{2}}{2} (v_{1} - v_{2})^{2} = v_{1}^{2} + v_{2}^{2} - 2 v_{1} v_{2} = \frac{3}{2} v_{0}^{2} + \frac{v_{0}^{2}}{2}$

solving we get relative velocity between the two particles

$v_{1} - v_{2} = \sqrt{2 v_{0}}$

Question

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A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the n^th power of R. If the period of rotation of the particle is T, then:

$T \propto R^{n / 2}$

$T \propto R^{3 / 2} f o r a n y n$

$T \propto R^{\frac{n}{2} + 1}$

$T \propto R^{\frac{n + 1}{2}}$

Solution

D.

$T \propto R^{\frac{n + 1}{2}}$

$m ω^{2} R = F o r c e \propto \frac{1}{R^{n}} (F o r c e = \frac{m v^{2}}{R}) \Rightarrow ω^{2} \propto \frac{1}{R^{n + 1}} \Rightarrow ω \propto \frac{1}{R^{\frac{n + 1}{2}}} T i m e P e r i o d T = \frac{2 π}{ω} T i m e P e r i o d T \propto R^{\frac{n + 1}{2}}$

Question

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It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is p_d; while for its similar collision with carbon nucleus at rest, fractional loss of energy is p_c. The values of p_d and p_c are respectively :

(0,1)

(.89,.28)

(.28,.89)

(0,0)

Solution

B.

(.89,.28)

For collision of a neutron with deuterium:

Applying conservation of momentum:

mv + 0 = mv₁ + 2mv₂ .....(i)
v₂ -v₁ = v ...... (ii)

Therefore, Collision is elastic, e = 1

From equ (i) and equ (ii) v₁ = -v/3

$P_{d} = \frac{\frac{1}{2} m v^{2} - \frac{1}{2} m v_{1}^{2}}{\frac{1}{2} m v^{2}} = \frac{8}{9} = 0.89$

Now, for the collision of neutron with carbon nucleus

Applying conservation of momentum

mv + 0 = mv₁ + 12mv₂ ....; (iii)

v = v₂-v₁ ....(iv)

$v_{1} = - \frac{11}{13} v P_{c} = \frac{\frac{1}{2} m v^{2} - \frac{1}{2} m {(\frac{11}{13} v)}^{2}}{\frac{1}{2} m v^{2}} = \frac{48}{169} \approx 0.28$

Question

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All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.

Solution

C.

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All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.

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For Daily Free Study Material Join wiredfaculty

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JEE physics

A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then: T∝ Rn/2 T∝ R3/2 for any n T∝Rn2+1 T∝ Rn+12

It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively : (0,1) (.89,.28) (.28,.89) (0,0)

All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation