Science and Education
Q: A cylinder of mass M and radius R is resting on two corner edges A and B as shown in figure. The normal reaction at the edges A and B are : (Neglect friction)
(A) $N_A = \sqrt{2} N_B $
(B) $N_B = \sqrt{3} N_A $
(C) $N_A = \frac{M g}{2} $
(D) $N_B = \frac{2\sqrt{3} M g}{5} $
On Solving , $N_B = \sqrt{3} N_A $
$N_A = \frac{M g}{2} $
Q: Two blocks each of mass m lie on a smooth table. They are attached to two other masses as shown in the figure. The pulleys and strings are light. An object O is kept at rest on the table. The sides AB and CD of the two blocks are made reflecting. The acceleration of two images formed in those two reflecting surfaces w.r.t. each other
(A) 5g/6
(B) 5g/3
(C) g/3
(D) 17g/6
Acceleration of block CD $ = \frac{2 m g}{2 m + m} = \frac{2}{3}g $
Acceleration of image in mirror AB
= 2 acceleration of mirror
$= 2 (-\frac{3 g}{4}) = -\frac{3}{2}g $
Acceleration of image in mirror CD $ = 2 (\frac{2 g}{3}) = \frac{4 g}{3}$
Acceleration of the two image w.r.t. each other
$= \frac{4 g}{3} – (- \frac{3 g}{2}) = \frac{17 g}{6}$
Q: In the figure shown all the surface are smooth.All the blocks A, B and C are movable, x-axis is horizontal and
y-axis vertical as shown. Just after the system is released from the position as shown.
(A)Acceleration of ‘A’ relative to ground is in negative y-direction
(B) Acceleration of ‘A’ relative to B is in positive x-direction
(C) The horizontal acceleration of ‘B’ relative to ground is in negative x-direction.
(D) The acceleration of ‘B’ relative to ground along the inclined surface of ‘C’ is greater than g sinθ
Due to the component of normal exterted by C on B, it moves in negative x-direction.
The force acting vertically downward on block B are mg and NA (normal reaction due to block A). Hence the component of net force on block B along the inclined surface of Bis greater than mg sinθ . Therefore the acceleration of ‘B’ relative to ground directed along the inclined surface of ‘C’ is greater than g sinθ
Q: A bead of mass m is attached to one end of a spring of natural length R and spring constant $k = \frac{(\sqrt{3}+1)mg}{R}$ . The other end of the spring is fixed at point A on a smooth vertical ring of radius R as shown in figure. The normal reaction at B just after it is released to
(a) $\displaystyle \frac{m g}{2} $
(b) $\displaystyle \sqrt{3} m g $
(c) $\displaystyle 3 \sqrt{3} m g $
(d) $\displaystyle \frac{3 \sqrt{3} m g}{2} $
Sol: The extension is spring is x = 2R cos 30° – R
x = (√3 – 1 )R
Applying Newton’s second law to the bead normal to circular ring at point B
N = k (√3 – 1 )R cos 30° + m g cos 30°
N = $k = \frac{(\sqrt{3}+1)mg}{R}$ (√3 – 1 )R cos 30° + m g cos 30°
$\displaystyle N = \frac{3 \sqrt{3} m g}{2} $