A light string passes over a frictionless pulley. To one of its ends a mass of 6 kg is attached. To its other end a mass of 10 kg is attached. The tension (in newton) in the thread will be [. Take g=9.8 m / s 2 ]

Question

A light string passes over a frictionless pulley. To one of its ends a mass of 6 kg is attached. To its other end a mass of 10 kg is attached. The tension (in newton) in the thread will be [. Take g=9.8 m / s 2 ] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

T =(2 m 1 m 2/ m 1+ m 2) g =(2 × 10 × 6/10+6) × 9.8=73.5 N

Q. A light string passes over a frictionless pulley. To one of its ends a mass of $6 k g$ is attached. To its other end a mass of $10 k g$ is attached. The tension (in newton) in the thread will be $[$ Take $g = 9.8 m / s^{2}$ ]

$T = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g$
$= \frac{2 \times 10 \times 6}{10 + 6} \times 9.8 = 73.5 N$

Q. A light string passes over a frictionless pulley. To one of its ends a mass of 6kg is attached. To its other end a mass of 10kg is attached. The tension (in newton) in the thread will be [ Take g=9.8m/s2 ]

T=m1​+m2​2m1​m2​​g =10+62×10×6​×9.8=73.5N

Q. A light string passes over a frictionless pulley. To one of its ends a mass of $6 k g$ is attached. To its other end a mass of $10 k g$ is attached. The tension (in newton) in the thread will be $[$ Take $g = 9.8 m / s^{2}$ ]

$T = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g$
$= \frac{2 \times 10 \times 6}{10 + 6} \times 9.8 = 73.5 N$