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HomeNEET UGphysicsc11phy_wep_pe_conserve › Per NCERT §5.8 eqn 5.9, a conservative force F(x…

Per NCERT §5.8 eqn 5.9, a conservative force F(x) is related to its potential V(x) by which relation?

AF(x) = ∫ V dx
BF(x) = + dV/dx
CF(x) = V(x)
DF(x) = − dV/dx
Answer & Solution
Correct answer: D. F(x) = − dV/dx
NCERT eqn 5.9 (1D): F(x) = − dV/dx. Negative gradient of PE gives the conservative force.
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