Physics · Mechanical Properties Of Fluids · NEET
The restoring force for a pendulum depends on the net downward pull on the bob. In air, this is the full weight m*g. Inside a liquid, buoyancy pushes the bob up, so the net downward pull becomes smaller: m*g minus the buoyant force. A weaker pull means the bob accelerates back more slowly, so each swing takes longer. Since T is inversely related to the square root of effective gravity, a smaller g_eff gives a bigger T. So the pendulum always swings slower in a liquid (as long as it stays fully submerged and the liquid is less dense than the bob).
Let the bob have volume V, density rho, so mass m = rho*V and weight = rho*V*g. The buoyant force from a liquid of density sigma is sigma*V*g (weight of displaced liquid). The net downward force = rho*V*g - sigma*V*g = V*g(rho - sigma). Effective acceleration g_eff = net force / mass = V*g(rho - sigma) / (rho*V) = g(1 - sigma/rho). Notice V cancels out, so the size of the bob does not matter.
The normal period is T = 2*pi*sqrt(L/g). In the liquid, replace g with g_eff: T' = 2*pi*sqrt(L/g_eff) = 2*pi*sqrt(L / [g(1 - sigma/rho)]). Divide the two: T'/T = sqrt(g / g_eff) = 1 / sqrt(1 - sigma/rho). So T' = T / sqrt(1 - sigma/rho). Because (1 - sigma/rho) is less than 1, the square root is less than 1, so T' is greater than T.
If sigma = rho, then g_eff = 0, so the buoyant force exactly balances the weight. There is no restoring force, so the pendulum will not oscillate and T becomes infinite. If sigma is greater than rho, the bob is lighter than the liquid it displaces, so it floats up instead of swinging as a pendulum. So this formula is valid only when the bob is denser than the liquid (sigma < rho).
In the standard NEET problem we ignore viscosity and assume an ideal (non-viscous) liquid. Then only buoyancy changes the time period, and the amplitude stays constant. In a real viscous liquid there is also a drag force that slowly reduces the amplitude (damped motion), but for the time period formula you use the ideal buoyancy result unless the question clearly asks about damping.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. As long as the bob is denser than water and fully submerged, buoyancy lowers the effective gravity, so the time period increases and the pendulum swings slower than in air.
T' = T / sqrt(1 - sigma/rho), where T is the period in air, sigma is the liquid density, and rho is the bob density. Equivalently T' = 2*pi*sqrt(L / [g(1 - sigma/rho)]).
Both the weight and the buoyant force are proportional to the bob's volume, so the volume cancels when you compute effective acceleration. The time period of a simple pendulum depends only on length and effective gravity, never on mass.
Then sigma is greater than rho, the bob floats up, and it no longer behaves as a swinging pendulum. When sigma equals rho, effective gravity is zero and the period becomes infinite (no oscillation).
The idea is the same: anything that changes the effective gravity g_eff changes the period. In a lift, g_eff depends on the lift's acceleration. In a liquid, g_eff = g(1 - sigma/rho) because of buoyancy. In both cases you just replace g with g_eff in T = 2*pi*sqrt(L/g_eff).
