A body floats in a liquid contained in a beaker. The whole system as shown in Figure falls freely under gravity. The upthrust on the body is
The whole system falls freely under gravity Upthrust = weight of fluid displaced
= (mass of fluid displaced) × g
For a freely falling body, g = 0
∴ Upthrust = 0.
The spring balance A reads 2 kg with a block m suspended from it. A balance B reads 5 kg when a beaker with liquid is put on the pan of the balance. The two balances are now so arranged that the hanging mass is inside the liquid in the beaker as shown in the figure. In this situation:
When the block of mass m is arranged as shown in the figure, an upthrust F_{T} will act on the mass which will decrease the reading on A.
Note :
According to Newton's third law, to each and every action, there is equal and opposite reaction.
So F_{T} will act on the liquid of the beaker which will increase the reading in B.
A vessel contains oil (density = 0.8 gm/cm^{3}) over mercury (density = 13.6 gm cm^{3}). A homogeneous sphere floats with half its volume immersed in mercury and the other half in oil.
The density of the material of the sphere in gm/cm^{3} is
Weight of sphere = Upthrust due to Hg + Upthrust due to oil
Two rods of different materials having coefficients of thermal expansion α_{1}, α_{2} and Young’s modulii Y_{1}, Y_{2} respectively are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of the rods. If α_{1} : α_{2} = 2 : 3, the thermal stresses developed in the two rods are equal provided Y_{1} : Y_{2} is equal to
Stress = Y × strain ∴ Stress = Yα ΔT
For first rod stress = Y_{1}α_{1}Δ T
For second rod stress = Y_{2}α_{2}ΔT
Since, stresses are equal
Water from a tap emerges vertically downwards with an initial spped of 1.0 m s^{–1}. The crosssectional area of the tap is 10^{–4} m^{2}. Assume that the pressure is constant throughout the stream of water, and that the flow is steady. The crosssectional area of the stream 0.15 m below the tap is
KEY CONCEPT :
The equation of continuity is : v_{1}A_{1} = v_{2}A_{2} where v and A represent the speed of water stream and its area of cross section, respectively. We are given that
v_{1} = 1.0 ms^{–1}
A_{1} = 10^{–4} m^{2}
v_{2} = velocity of water stream at 0.15 m below the tap A_{2} = ?
For calculating v_{2}
u = 1 m/s; s = 1.5 m, a = 10 m/s^{2} and v = ?
v^{2} – u^{2} = 2as
v^{2} – 1 = 2 × 10 × 0.15
⇒ v = 2 m/s
A solid sphere of radius R and density ρ is attached to one end of a massless spring of force constant k. The other end of the spring is connected to another solid sphere of radius R and density 3ρ. The complete arrangement is placed in a liquid of density 2ρ and is allowed to reach equilibrium. The correct statement(s) is (are)
Consider the equilibrium of the system of both spheres and the spring.
The weight of system
This is to be balanced by the buoyant force.
This can be possible only when the light sphere is completely submerged. In this way the buoyant force
Now considering the equilibrium of the heavy sphere Fs + B = W
∴ Fs = W – B
In plotting stress versus strain curves for two materials P and Q,a student by mistake puts strain on the yaxis and stress on the xaxis as shown in the figure. Then the correct statement(s) is (are)
The maximum stress that P can withstand before breaking is greater than Q. Therefore (A) is a correct option.
The strain of P is more than Q therefore P is more ductile. Therefore (B) is a correct option.
For a given strain, stress is more for Q. Therefore Y_{Q} > Y_{P}.
A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own gravity. If P(r) is the pressure at r(r < R), then the correct option(s) is (are)
Let us consider an elemental mass dm shown in the shaded portion.
Here
B and C are correct options.
Two spheres P and Q of equal radii have densities ρ_{1} and ρ_{2}, respectively. The spheres are connected by a massless string and placed in liquids L_{1} and L_{2} of densities σ_{1} and σ_{2} and viscosities η_{1} and η_{2}, respectively. They float in equilibrium with the sphere P in L_{1} and sphere Q in L_{2} and the string being taut (see figure). If sphere P alone in L_{2} has terminal velocity alone in L_{1} has terminal velocity
From the figure it is clear that
(a) σ_{2} > σ_{1}
(b) ρ_{2} > σ_{2} [As the string is taut]
(c) ρ_{1} < σ_{1} [As the string is taut]
∴ ρ_{1} < σ_{1} < σ_{2} < ρ_{2}
When P alone is in L2
is negative as ρ_{1} < σ2
Where r is radius of sphere.
When Q alone is in L_{1}
is positive as ρ_{2} > σ_{1}
Therefore option (d) is correct
...(i)
For equilibrium of Q
...(ii)
For equilibrium of P
...(iii)
(iii) – (ii) gives
ρ_{1} – σ_{2} = σ_{1} – ρ_{2} ...(iv)
From (i) and (iv)
∴ A is also a correct option
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