A 5 kg brick of 20 cm × 10 cm × 8 cm dimensionless lying on the largest base. It is now made to stand with length vertical. If g=10 ms -2, then the amount of work done is

Question

A 5 kg brick of 20 cm × 10 cm × 8 cm dimensionless lying on the largest base. It is now made to stand with length vertical. If g=10 ms -2, then the amount of work done is (A) 3 J (B) 5 J (C) 7 J (D) 9 J

Tardigrade Admin · Accepted Answer

Initial height of CG =4 cm Final height of CG =10 cm Increase in height =6 cm =0.06 m Work done =5 × 10 × 0.06=3 J

Q. A $5 k g$ brick of $20 c m \times 10 c m \times 8 c m$ dimensionless lying on the largest base. It is now made to stand with length vertical. If $g = 10 m s^{- 2}$ , then the amount of work done is

3 J

5 J

7 J

9 J

Initial height of $CG = 4 c m$
Final height of $CG = 10 c m$
Increase in height $= 6 c m = 0.06 m$
Work done $= 5 \times 10 \times 0.06 = 3 J$

Q. A 5kg brick of 20cm×10cm×8cm dimensionless lying on the largest base. It is now made to stand with length vertical. If g=10ms−2, then the amount of work done is

3 J

5 J

7 J

9 J

Initial height of CG=4cm Final height of CG=10cm Increase in height =6cm=0.06m Work done =5×10×0.06=3J

Q. A $5 k g$ brick of $20 c m \times 10 c m \times 8 c m$ dimensionless lying on the largest base. It is now made to stand with length vertical. If $g = 10 m s^{- 2}$ , then the amount of work done is

Initial height of $CG = 4 c m$
Final height of $CG = 10 c m$
Increase in height $= 6 c m = 0.06 m$
Work done $= 5 \times 10 \times 0.06 = 3 J$