A particle of mass m=5kg is moving with a uniform speed 3√2m/s in XY plane along the line y=x+4. The angular momentum about the origin is [in kg m2/s]
A
40
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
60
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
zero
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
20
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B60
Given, Mass of particle m=5kg Speed of particle v=3√2m/s As we know, L=mvr⊥ where r⊥ is the shortest /perpendicular distance, of the particle from the origin. As we know, Perpendicular distance between a point and a line is given by, r⊥=∣∣∣Ax1+Bx2+C√A2+B2∣∣∣=∣∣
∣
∣∣x−y+4√12+(−1)2∣∣
∣
∣∣=∣∣∣0−0+4√2∣∣∣ (substituting (x,y)=(0,0) for origin) r⊥=2√2