A55 g mouse runs out to the end of the 17 cm-long minute hand of a grandfather clock when the clock reads 10 minutes past the hour. what torque does the mouse's weight exert about the rotation axis of the clock hand?

The equation for torque is written below:

τ = rFsinθ
where
τ is torque 
r is the length of minute hand
F is the force exerted by the mouse
θ is the angle

Every number interval in the clock has an angle of 30°. Because 10 minutes past the hour coincides with the number 10, θ = 30°*2 = 60°. For the force, it is simply the weight of the mouse which is
F = mg = (55g)*(1 kg/1000g)*(9.81 m/s²) = 0.53955 N

Substituting the rest of the values,

τ = (17cm)*(1 m/100 cm)*(0.53955 N)*(sin60°)
τ = 0.079 N·m