Three point particles of masses 1.0 kg, 1.5 kg and 2.5 kg are placed at three corners of a right angle triangle of sides 4.0 cm, 3.0 cm and 5.0 cm as shown in the figure. The center of mass of system is at a point |
[JEE MAIN Held on 07-01-2020 Morning] |
A) 1.5 cm right and 1.2 cm above 1 kg mass
B) 2.0 cm right and 0.9 cm above 1 kg mass
C) 0.9 cm right and 2.0 cm above 1 kg mass
D) 0.6 cm right and 2.0 cm above 1 kg mass
Correct Answer: C
Solution :
[c] |
\[{{X}_{cm}}=\frac{1\times 0+1.5\times 3+2.5\times 0}{1+1.5+2.5}=\frac{1.5\times 3}{5}=0.9\,cm\] |
\[{{Y}_{cm}}=\frac{1\times 0+1.5\times 0+2.5\times 4}{1+1.5+2.5}=\frac{2.5\times 4}{5}=2\,cm\] |
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