You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Newton's Second Law: Rotatory Version

A rod of leng...

Question

A rod of length of 1 m is attached to the string at its mid point. A 6 kg mass is attached to end of the rod and another mass M is hung at a distance of 30 cm from the center of the rod as shown in the figure. Assume mass of rod is negligible. Find out the value of M.

3 kg

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 kg

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6 kg

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10 kg

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

30 kg

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D 10 kg
Given : $r_{1} = 30$ cm $r_{2} = 50$ cm $m_{2} = 6 k g$
As the rod is in equilibrium, thus net torque about the point O is zero i.e $τ_{o} = 0$
$∴$ $(m_{1} g) r_{1} - (m_{2} g) r_{2} = 0$
OR $m_{1} r_{1} = m_{2} r_{2}$
OR $M \times 30 = 6 \times 50$ $⟹ M = 10 k g$

Suggest Corrections

Similar questions

Q. A uniform rod of mass

2 kg

is hanging from a thread attached at the midpoint

(O)

of the rod. A block of mass

m = 6 kg

hangs at the left end of the rod and a block of mass

M

hangs at the right end at a distance of

30 cm

from the mid point

(O)

. If the system is in equilibrium, calculate the mass

(M)

, given that overall length of the rod is

100 cm

. Take

g = 10 {m/s}^{2}

Q. A uniform rod of mass

2 kg

is hanging from a thread attached at the midpoint

(O)

of the rod. A block of mass

m = 6 kg

hangs at the left end of the rod and a block of mass

M

hangs at the right end at a distance of

30 cm

from the mid point

(O)

. If the system is in equilibrium, calculate the mass

(M)

, given that overall length of the rod is

100 cm

. Take

g = 10 {m/s}^{2}

Shown in the figure is a rigid and uniform one meter long rod

A B

held in horizontal position by two strings tied to its ends and attached to the ceiling. The rod is of mass

m

and has another weight of mass

2 m

hung at a distance of

75 cm

from

A

. The tension in the string at

A

is :

Q. A rod of negligible mass having length

l = 2 m

is pivoted at its center and two masses,

6 kg

and

4 kg

are hung from the ends as shown in the figure. Find the initial angular acceleration of the rod if it is suddenly released from the horizontal position.

Take

g = 10 {m/s}^{2}

Q. Two particles of mass

5 kg

and

10 kg

respectively are attached to the two ends of a rigid rod of length

1 m

with negligible mass. The centre of mass of the system from the

5 kg

particle is nearly at a distance of :

Join BYJU'S Learning Program

A rod of length of 1 m is attached to the string at its mid point. A 6 kg mass is attached to end of the rod and another mass M is hung at a distance of 30 cm from the center of the rod as shown in the figure. Assume mass of rod is negligible. Find out the value of M.

The correct option is D 10 kgGiven : r1=30 cm r2=50 cm m2=6kgAs the rod is in equilibrium, thus net torque about the point O is zero i.e τo=0∴ (m1g)r1−(m2g)r2=0OR m1r1=m2r2OR M×30=6×50 ⟹M=10kg

The correct option is D 10 kg
Given : $r_{1} = 30$ cm $r_{2} = 50$ cm $m_{2} = 6 k g$
As the rod is in equilibrium, thus net torque about the point O is zero i.e $τ_{o} = 0$
$∴$ $(m_{1} g) r_{1} - (m_{2} g) r_{2} = 0$
OR $m_{1} r_{1} = m_{2} r_{2}$
OR $M \times 30 = 6 \times 50$ $⟹ M = 10 k g$