Skip to content
dronacharya.ai
← JEE Advanced 2021 · Paper 1
PhysicsMultiple correct+4 / -2Question 16 of 57

JEE Advanced 2021 · Paper 1, Physics question 16

A cylindrical tube, with its base as shown in the figure, is filled with water. It is moving down with a constant acceleration aa along a fixed inclined plane with angle θ=45∘\theta = 45^\circ. P1P_1 and P2P_2 are pressures at points 1 and 2, respectively, located at the base of the tube. Let β=(P1−P2)/(ρgd)\beta = (P_1 - P_2)/(\rho g d), where ρ\rho is density of water, dd is the inner diameter of the tube and gg is the acceleration due to gravity. Which of the following statement(s) is(are) correct?

A vertical cylindrical tube of inner diameter d, filled with water, whose base is cut along a fixed 45° incline and rests on it. Point 2 is the upper end of the slanted base and point 1 the lower end. The tube slides down the incline with acceleration a (arrow along the incline, downward); gravity g points vertically down. The water surface is drawn tilted.
Choose one or more options
Answer and worked solution

Official answer: A, C

Effective gravity. In the frame of the tube, the water feels g⃗eff=g⃗−a⃗\vec g_{eff}=\vec g-\vec a. Take xx horizontal (downhill side) and yy up:
g⃗=(0,−g),a⃗=a2(1,−1),g⃗eff=(−a2, −g+a2).\vec g=(0,-g),\qquad\vec a=\frac a{\sqrt2}(1,-1),\qquad\vec g_{eff}=\left(-\frac a{\sqrt2},\,-g+\frac a{\sqrt2}\right).

Pressure difference. Pressure increases along g⃗eff\vec g_{eff}. On the 45∘45^\circ base, point 1 is dd further downhill and dd lower than point 2:
P1−P2=ρ g⃗eff⋅(d,−d)=ρd(g−2a).P_1-P_2=\rho\,\vec g_{eff}\cdot(d,-d)=\rho d\left(g-\sqrt2a\right).
So β=1−2ag\beta=1-\dfrac{\sqrt2a}g.

Check each option.

  • a=g2a=\frac g{\sqrt2}: β=0\beta=0. (A) is true and (B) is false.
  • a=g2a=\frac g2: β=1−12=2−12\beta=1-\frac1{\sqrt2}=\frac{\sqrt2-1}{\sqrt2}. (C) is true and (D) is false.

(A), (C)

Topics:Pressure and Pascal's lawPseudo forces and non-inertial frames
← Previous questionNext question →