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- Question
If 2x – y + 1 = 0 is a tangent to the hyperbola x2/a2 – y2/16 = 1, then which of the following cannot be the sides of a right-angled triangle?
Options- A. a, 4, 1
- B. a, 4, 2
- C. 2a, 8, 1
- D. 2a, 4, 1
- Correct Answer
- A, B, C
- 1. An ideal gas is expanded from (p1, V1, T1) to (p2, V2, T2) under different conditions. The correct statement(s) among the following is(are)
Options- A. The work done on the gas is maximum when it is compressed irreversibly from (p2, V2) to (p1, V1) against
constant pressure p1
- B. If the expansion is carried out freely, it is simultaneously both isothermal as well as adiabatic
- C. The work done by the gas is less when it is expanded reversibly from V1 to V2 under adiabatic conditions as compared to that when expanded reversibly from V1 to V2 under isothermal conditions
- D. The change in internal energy of the gas is (i) zero, if it is expanded reversibly with T1= T2, and (ii) positive, if it is expanded reversibly under adiabatic conditions with T1¹ T2 Discuss
- 2. A flat plate is moving normal to its plane through a gas under the action of a constant force F. The gas is kept at very low pressure. The speed of the plate v is much less than the average speed u of the gas molecules. Which of the following options is/are true?
Options- A. The pressure difference between the leading and trailing faces of the plate is proportional to uv
- B. resistive force experienced by the plate is proportional to v
- C. The plate will continue to move with constant non-zero acceleration, at all times
- D. At a later time the external force F balances the resistive force Discuss
Mathematics problems
Search Results
Correct Answer: A, B, C
Correct Answer: A, B, D
Explanation:
n = number of molecules per unit volume
u = average speed of gas molecules
When the plate is moving with speed v, the relative speed of molecules w.r.t. leading force = v + u
Coming head-on, momentum transformed to plate per collision = 2m(u + v)
Number of collision in time: Δt = ((v + v0)nΔtA)/2
where A = Surface area
So, momentum transferred in time Δt = m(u + v)2nAΔt from front surface
Similarly momentum transferred in time = m(u – v)2nAΔt from back surface
Net force = mnA[(v + u)2 – (u – v)2]
= mnA[4vu]
F ∝ v
Pleading– Ptrailing = mn[u + v]2 – mn[u – v]2
= mn[4uv]
= 4mnuv
ΔP ∝ uv
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