A metal ball immersed in alcohol weight W1 at 0°C and W2 at 50°C. the coefficient of cubical expansion…

Q: A metal ball immersed in alcohol weight W1 at 0°C and W2 at 50°C. the coefficient of cubical expansion of the metal is less than that of the alcohol. Assuming that the density of the metal is large compared to that of alcohol, it can be shown that

(a) W1 > W2

(b) W1 = W2

(c)W1 < W2

(d) All of these

Ans: (c)

Sol: Let Vs = volume of solid & ρL = density of liquid

$\large W_{apparent} = W_{actual} – Upthrust $

Upthrust F = Vs ρL g

At higher temperature ,

Upthrust F’ = Vs‘ ρL‘ g

$\large \frac{F’}{F} = \frac{V_s’}{V_s} .\frac{\rho_L’}{\rho_L} = \frac{1+\gamma_s \Delta \theta}{1+\gamma_L \Delta \theta}$

Since , $\large \gamma_s < \gamma_L $

Hence, $\large F’ < F $

Therefore , $\large W’_{app} > W_{app} $

$\large W_2 > W_1 $

A particle moves with simple harmonic motion is a straight line. In first τ sec, after starting from rest…

Q: A particle moves with simple harmonic motion is a straight line. In first τ sec, after starting from rest it travels a distance ‘ a ‘ and in next τ sec, it travels 2a , in same direction, then

(a)Amplitude of motion is 3a

(b)Time period of oscillations is 8π

(c)Amplitude of motion is 4a

(d)Time period of oscillations is 6π

Ans: (d)

Sol: As particle starting from rest .

x = A cosωt

At t= 0 ; x = A

When t = τ ; x = A – a

When t = 2τ ; x = A – 3a

Therefore ,

A – a = A cosωτ

A – 3a = A cos2ωτ

using formula ;

cos2ωτ = 2cos2ωτ – 1

$\large \frac{A-3a}{A} = 2 (\frac{A-a}{A})^2 – 1$

on solving this we get

A = 2a

2a – a = 2a cosωτ

cosωτ = 1/2

ωτ = π/3

$\large \frac{2\pi}{T}\tau = \frac{\pi}{3}$

A small block is connected to one end of a massless springs of unstretched length 4.9 m…

Q: A small block is connected to one end of a massless springs of unstretched length 4.9 m. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by 0.2 m and released from rest at t = 0. It then executes simple harmonic motion with angular frequency ω = π/3 rad/s. simultaneously at t=0, a small pebble is projected with speed v from point P at an angle of 45° as shown in the figure. Point P is at a horizontal distance of 10m from O. if the pebble hits the block at t=1s, the value of v is
(take g=10 m/s2)

Q & A

(a) $ \sqrt{50}$ m/s

(b) $ \sqrt{51}$ m/s

(c) $ \sqrt{52}$ m/s

(d) $ \sqrt{53}$ m/s

Ans: (a)

Sol: time of flight t = 1 sec

$\large \frac{2v sin\theta}{g} = 1$

$\large \frac{2 v sin45^o}{10} = 1$

$v = \sqrt{50}$ m/s

A point mass is subjected to two simultaneous sinusoidal displacements in x-direction…

Q: A point mass is subjected to two simultaneous sinusoidal displacements in x-direction, x1(t)= A sin ωt and x2(t) = A sin (ωt+2π/3). Adding a third sinusoial displacement x3(t) = B sin (ωt+ φ) brings the mass to a complete rest. The values of B and φ are

(a) $\large \sqrt{2} A , \frac{3 \pi}{4}$

(b) $\large A , \frac{4 \pi}{3}$

(c) $\large \sqrt{3} A , \frac{5 \pi}{6}$

(d) $\large A , \frac{ \pi}{3}$

Ans: (b)

A wooden block perform SHM on a frictionless surface with frequency ν0. The block carries a charge +Q on its surface…

Q: A wooden block perform SHM on a frictionless surface with frequency ν0. The block carries a charge +Q on its surface. If now a uniform electric field $\vec{E}$ is switched-on as shown, then the SHM of the block will be

Q & A

(a)Of the same frequency and with shifted mean position

(b)Of the same frequency and with same mean position

(c)Of the changed frequency and with shifted mean position

(d)Of the changed frequency and with same mean position

Ans: (a)

Sol: As constant external force can only change the mean position . When electric field is switched on , mean position will be obtained after a compression of x0

$\large k x_0 = Q E$

$\large x_0 = \frac{Q E}{k}$