A $8 k g$ metal block of dimension $16 c m \times 8 c m \times 6 c m$ is lying on a table with its face of largest area touching the table. If $g = 10 m s^{- 2}$ , the minimum amount of work done in making it stand with its length vertical is :

0.4 J

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

6.4 J

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

4 J

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

12.8 J

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

Open in App

Solution

The correct option is C $4 J$
$M = 8 k g$
$d i m e n s i o n = 16 \times 8 \times 6$
Face with largest area $= 16 \times 8$
Initial center of mass height $\frac{6 c m}{2} = 3 c m = 0.03$
Work done from bringing centre of mass from $3 c m$ to $8 c m$
$\frac{16}{2} = 8$ (angle side)
$= 0.08 m$
$w = m g (h_{2} - h_{1})$
$8 \times 10 (0.08 - 0.03)$
$4 J$

Suggest Corrections

Similar questions

8 kg

16 cm \times 8 cm \times 6 cm

g = 10 {ms}^{- 2},

A rod of mass m and length l is lying on a horizontal table. The work done in making it stand vertically on one end will be

m

l

6 m \times 4 m \times 2 m

1.5 g m / c . c

(g = 10 m s^{- 1})

A rod of mass m and length l is lying on a horizontal table. The work done in making it stand vertically on one end will be

Join BYJU'S Learning Program