Two masses m1 and m2 are attached to a spring balance S as shown in Figure. If m1 > m2 then the reading of spring balance will be

Fnet = ma ,From FBD of m1 , m1g - T1 = m1aFrom FBD of m2 , T2 - m2g = m2a take T1 = T2 ,solving the above eq’s, we get ‘ a ’

Two masses m1 and m2 are attached to a spring balance S as shown in Figure. If m1 > m2 then the reading of spring balance will be

Fnet = ma ,From FBD of m1 , m1g - T1 = m1aFrom FBD of m2 , T2 - m2g = m2a take T1 = T2 ,solving the above eq’s, we get ‘ a ’

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Two masses m₁ and m₂ are attached to a spring balance S as shown in Figure. If m₁ > m₂ then the reading of spring balance will be

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(m₁ + m₂)

$\frac{2 m_{1} m_{2}}{m_{1} + m_{2}}$

$\frac{m_{1} m_{2}}{m_{1} + m_{2}}$

(m₁ - m₂)

answer is C.

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Fnet = ma ,From FBD of m₁ , m₁g - T₁ = m₁a
From FBD of m₂ , T₂ - m₂g = m₂a take T₁ = T₂ ,solving the above eq’s, we get ‘ a ’

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Two masses m1 and m2 are attached to a spring balance S as shown in Figure. If m1 > m2 then the reading of spring balance will be

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Two masses m₁ and m₂ are attached to a spring balance S as shown in Figure. If m₁ > m₂ then the reading of spring balance will be