# 409 - Physics Questions Answers

**Asked By: KRISHAV RAJ SINGH**

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**Solution by Joshi sir**

$\frac{k{q}_{1}{q}_{2}}{{{l}_{3}}^{2}}=\frac{k{q}_{2}{q}_{3}}{{{l}_{1}}^{2}}=\frac{k{q}_{3}{q}_{1}}{{{l}_{2}}^{2}}\phantom{\rule{0ex}{0ex}}{l}_{1}+{l}_{2}+{l}_{3}=L\phantom{\rule{0ex}{0ex}}fromthese2equations\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{L\frac{1}{\sqrt{{q}_{1}}}}{\frac{1}{\sqrt{{q}_{1}}}+\frac{1}{\sqrt{{q}_{2}}}+\frac{1}{\sqrt{{q}_{3}}}},\frac{L\frac{1}{\sqrt{{q}_{2}}}}{\frac{1}{\sqrt{{q}_{1}}}+\frac{1}{\sqrt{{q}_{2}}}+\frac{1}{\sqrt{{q}_{3}}}},\frac{L\frac{1}{\sqrt{{q}_{3}}}}{\frac{1}{\sqrt{{q}_{1}}}+\frac{1}{\sqrt{{q}_{2}}}+\frac{1}{\sqrt{{q}_{3}}}}\phantom{\rule{0ex}{0ex}}$

**Asked By: AYUSH PAINE**

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**Joshi sir comment**

${X}_{C}=50\Omega and{X}_{L}=100\Omega areinparallel.\phantom{\rule{0ex}{0ex}}Resul\mathrm{tan}tis100\Omega capaci\mathrm{tan}ce.\phantom{\rule{0ex}{0ex}}Now100\Omega capaci\mathrm{tan}ceand100\Omega induc\mathrm{tan}cegivereac\mathrm{tan}ce0\Omega \phantom{\rule{0ex}{0ex}}andresis\mathrm{tan}ceis100\Omega .SoonlyoptionDiscorrect.$

A rotating ball hits a rough horizontal plane with a vertical velocity and angular velocity . Given that the coefficient of friction is and the vertical component of the velocity after the collision is , find:

a) the angular velocity after collision;

b) the impulsive ground reaction during the collision

**Asked By: UDDESH KUMAR SABAT**

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**Joshi sir comment**

on applying momentum impulse theorem along vertical

-mv+Ndt = mv/2 (1)

by angular momentum angular impulse theorem

$2m\omega {r}^{2}/5-\mu Ndtr=2m{\omega}^{,}{r}^{2}/5\left(2\right)$By eq. (2) get angular velocity.

impulsive ground reaction = $\sqrt{{\left(Ndt\right)}^{2}+{\left(\mu Ndt\right)}^{2}}$by using eq. (1), solve it

A particle a having a charge of 5.0×10

is fixed in. a vertical wall. A second particle B of mass 100 g and. having equal charge is supended by a silk thread. of length 30 cm form the wall. The point of suspension is. 30 cm above the particle A. Find the angle of the thread. with the vertical when it stays in equilibrium.

**Asked By: UDDESH KUMAR SABAT**

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**Solution by Joshi sir**

a toy car of mass 500g travels with a uniform velocity of 25m/s for 5 seconds, the brakes are then applied and the car is uniformly retarded and comes to rest in further 10s calculate the retardation

**Asked By: SARABJEET**

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**Joshi sir comment**

use v = u + at

u = 25, v = 0 and t = 10

Hi please slove these question also...

A ) **Two particles of mass 3 kg and 4 kg are connected by a light inelastic string passing over a smooth fixed pulley. The system is released from rest with the string taut and both particles at a height of 2 m above the ground. Find the velocity of the 3 kg mass when the 4 kg mass reaches the ground, and find when the 4 kg mass reaches the ground. **

**B) Briefly describe an experiment to find the coefficient of friction between brick and tile. You may assume access to a tile slab, a number of bricks and basic scientific equipment, such as a pulley, weighing machine, etc.**

Please reply ASAP...Waiting for your reply.

**Asked By: RAAJ**

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**Joshi sir comment**

(A)4g-T = 4a

T-3g = 3a

solve for a

use h = 0t+1/2at^{2} for getting t

and v = 0 +at for v

(B) get F applied and acceleration of the body

then use F-kmg = ma

for getting k (coefficient of friction)

Please slove the below mention question:

A) A uniform ladder of weight w rests on rough horizontal ground against a smooth vertical wall. The vertical plane containing the ladder is perpendicular to the wall and the ladder is inclined at an angle θ to the *vertical. *Prove that, if the ladder is on the point of slipping and µ is the coefficient of friction between it and the ground, then tanθ = 2µ

B) Two particles of mass 3 kg and 4 kg are connected by a light inelastic string passing over a smooth fixed pulley. The system is released from rest with the string taut and both particles at a height of 2 m above the ground. Find the velocity of the 3 kg mass when the 4 kg mass reaches the ground, and find when the 4 kg mass reaches the ground.

C) A box of mass 14 kg is placed in the back of a van. The coefficient of friction between the box and the floor is 0.5. What happens to the box if the lorry moves off with an acceleration of

(a) 4 ms^{-2}

(b) 5 ms^{-2}

(c) 8 ms^{-2}

(Take g = 10 ms^{-2})

Can i get answer before 3pm today. Please i request you to help me out from the issue...

**Asked By: RAAJ**

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**Joshi sir comment**

(a)let R and N are reactions by the ground and wall then for equilibrium

μR = N

R = w

and about ground point

N*lcosθ = w*lsinθ/2

now solve

(b) 4g-T = 4a

T-3g = 3a

solve for a

use h = 0t+1/2at^{2} for getting t

and v = 0 +at for v

(c) for a = 8 pseudo force = 14*8 and friction max = 0.5*14*10 so box will fall in backward direction

A 50 kg gymnast falls freely from a height of 4 m on to a trampoline. The trampoline then bounces her back upward with a speed equal to the speed at which she first struck the trampoline. What is the average force the trampoline applies on the gymnast ?

A) 50 N B) 200 N C) 500 N D) 2000 N E) more information is required

**Asked By: VAIBHAV GUPTA**

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**Joshi sir comment**

rain is falling down vertically 2m/s. a person is walking with velocity 2m/s on horizantal road. the angle made by the umbrella with vertical to protect from rain is

**Asked By: VISHISTA**

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**Joshi sir comment**