Pseudo Force

Question 1:

A block is kept on the frictionless inclined surface with angle of inclination . The incline is given an acceleration a to keep the block stationary w.r.t incline plane. Then a is equal to

1. g/tan α

2. g cosec α

3. g

4. g tan α

View Answer

Taking incline plane as frame of reference, Pseudo force ma act towards right.

component of Pseudo force up the incline is m.a.cos α which is balanced by m.g.sin α. so,

m.a.cos α = m.g.sin α

⇒a= g tan α

Question 2:

A trolley of mass 8 kg is standing on a frictionless surface inside which an object of mass 2 kg is suspended. A constant force F starts acting on the trolley as a result of which the string stood at an angle of 37⁰ from the vertical (bob at rest relative to trolley) Then:

1. acceleration of the trolley is 40/3 m/sec.sec

2. force applied is 60 N

3. force applied is 75 N

4. force applied is 25 N

View Answer

Using concept of Pseudo force

a= g tanθ

⇒ a = g tan 37 = 10 ×3/4= 7.5 m/s^2

Using Equation of motion

⇒Fnet = 10×2.5 = 25 N

Question 3:

A man of weight W = Mg is standing on a lift which is moving upward with an acceleration a. If g is the acceleration due to gravity, the apparent weight of the man is

1. W (1 + a/g)

2. W (1 - a/g)

3. W

4. W(a+g)

View Answer

Taking lift as frame of reference, T = mg +ma = mg (1 +a/g) = W (1+ a/g)

Question 4:

A block of mass m is in contact with the cart. The coefficient of static friction between the block and the cart is ü. The acceleration a of the cart that prevent the block from falling will be

neet pseudo force questions

1. a> (mg/ü)

2. a> (g/ü.m)

3. a ≥ (g/ü)

4. a ≤ (g/ü)

View Answer

The condition to prevent the block from falling is that the friction force must be at least equal to the weight of the block:

$f_s \geq mg$

Since static friction is given by $f_s = \mu N$ and the normal force is due to pseudo force $N = ma$ , we get:

$\mu ma \geq mg$

Dividing both sides by $\mu m$ :

$a \geq \frac{g}{\mu}$

Thus, the required acceleration of the cart is $a \geq \frac{g}{\mu}$ .

Question 5:

A truck is stationary and has a bob suspended by a light string, in a frame attached to the truck. The truck suddenly moves to the right with an acceleration of a. The pendulum will tilt

1. To the left and angle of inclination of the pendulum with the vertical is tan-¹(g/a)

2. To the left and angle of inclination of the pendulum with the vertical is sin-¹(g/a)

3. To the left and angle of inclination of the pendulum with the vertical is tan-¹(a/g)

4. To the left and angle of inclination of the pendulum with the vertical is sin-¹(a/g)

View Answer

When the truck accelerates to the right with $a$ , a pseudo force $ma$ acts to the left on the bob in the truck's frame. The bob reaches equilibrium where the tension components balance forces:

$\tan \theta = \frac{\text{Pseudo force}}{\text{Weight}} = \frac{ma}{mg} = \frac{a}{g}$ $\theta = \tan^{-1} \left(\frac{a}{g} \right)$

Thus, the pendulum tilts to the left at an angle $\theta = \tan^{-1}(a/g)$ with the vertical.

Question 6:

A block of mass m is placed on a smooth inclined wedge ABC of inclination θ as shown in the figure. The wedge is given an acceleration a
towards the right. The relation between a and θ for the block to remain stationary on the wedge is:

Neet pseodo force questions

1. a= g / (cosec θ)

2. a = g/(sinθ)

3. a= g tanθ

4. a = g cosθ

View Answer

For the block to remain stationary relative to the wedge,

the component of pseudo force up the incline balances the gravitational component along down the incline:

$ma \cos \theta = mg \sin \theta$

$⇒ a = g \tan θ$