Rishabhiiitd/PhysicsQA huggingface dataset:
Topic: Friction
Q11: The minimum force required to start pushing a body up a rough (frictional coefficient μ ) inclined plane F₁ while the minimum force needed to prevent it from sliding down is F₂. If the inclined plane makes an angle θ from the horizontal such that tan θ=2 μ then the ratio F₁ / F₂ is
Options: (a) 4
(b) 1
(c) 2
(d) 3
Answer: Answer: (d) 3
Solution: Solution: We have to work against the sliding force due to gravity and frictional force to push upwards. Therefore, the force F₁=mg sin θ+μ mg cos θ.
To stop the body from sliding work is done against sliding force but frictional force stops the body from sliding.
Therefore, the force F₂=mg sin θ-μ mg cos θ.
F₁ / F₂=(mg sin θ+μ mg cos θ) /(mg sin θ-μ mg cos θ)
F₁ / F₂=(tan θ+μ) /(tan θ-μ)
(since tan θ=2 μ )
F₁ / F₂=(2 μ+μ) /(2 μ-μ)
F₁ / F₂=3
Topic: Friction
Q12: A body of mass m=10⁻² kg is moving in a medium and experiences a frictional force F=-kv². Its initial speed is v₀=10 ms⁻¹. If, after 10 s, its energy is 1 / 8 mv₀², the value of k will be
Options: (a) 10⁻³ kg m⁻¹
(b) 10⁻³ kg s⁻¹
(c) 10⁻⁴ kg m⁻¹
(d) 10⁻¹ kg m⁻¹ s⁻¹
Answer: Answer: (c) 10⁻⁴ kg m⁻¹
Solution: Solution: Mass m=10² Kg
Initial velocity v₀=10 ms⁻¹
Time t=10 s
Frictional Force F=-kv²
1 / 2 mv_f²=1 / 8 mv₀²
v_f=v₀ / 2
v_f=10 / 2=5 m / s
Now, the force is
(10⁻²) dv / dt=-kv²
∫_{10}^{5} (dv / v²)=-100 k ∫_{0}^{10} (dt)
1 / 5=1 / 10=100 kx 10
k=10⁻⁴
Topic: Friction
Q13: A block of mass m is placed on a surface with a vertical cross-section given by y=x³ / 6. If the coefficient of friction is 0.5 , the maximum height above the ground at which the block can be placed without slipping is
Options: (a) 1 / 3 m
(b) 1 / 2 m
(c) 1 / 6 m
(d) 2 / 3 m
Answer: Answer: (c) 1 / 6
Solution: Solution: For equilibrium under limiting friction
mg sin θ=μ mg cos θ
tan θ=μ
The equation of the surface is given by y=x³ / 6
Slope dy / dx=x² / 2
From (1) and (2) we get
μ=x² / 2=0.5
x=1
So, y=1 / 6
Topic: Thermal Expansion
Q2: The ratio of work done by an ideal monoatomic gas to the heat supplied to it in an isobaric process is
Options: (a) 2 / 5
(b) 3 / 2
(c) 3 / 5
(d) 2 / 3
Answer: (a) 2 / 5
Solution: For an ideal gas in an isobaric process, Heat supplied, Q=nCp ΔT Work done, W=P ΔV=nR ΔT Required Ratio =W / Q Required Ratio =(nR ΔT)/(nCp ΔT) R / Cp=R/(γR / γ-1) [since γ=5 / 3, for a monoatomic gas ] =2 / 5
Topic: Thermal Expansion
Q3: The work of 146 kJ is performed in order to compress one-kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7°C. The gas is (R=8.3 J mol−1 K−1)
Options: (a) monoatomic
(b) diatomic
(c) triatomic
(d) a mixture of monoatomic and diatomic
Answer: (b) diatomic
Solution: Increase in temperature, ΔT=T2−T1=7 K (Since it is change in temperature so its value will be same in Kelvin and degree celsius) Work done on the system, W=146 KJ=146 × 1000 J Number of moles of gas, n=1000 Work done in an adiabatic process, W=nR(T2−T1)/γ-1 put the values, 146000=1000 × 8.3 × 7 /(γ-1) γ-1=0.39 γ≈1.4 Hence the gas is diatomic
Topic: Thermal Expansion
Q4: Two moles of an ideal monatomic gas occupies a volume V at 27°C. The gas expands adiabatically to a volume 2 V. Calculate (i) the final temperature of the gas and (ii) change in its internal energy.
Options: (a) (i) 198 K (ii) 2.7 kJ
(b) (i) 195 K (ii) 2.7 kJ
(c) (i) 189 K (ii) 2.7 kJ
(d) (i) 195 K (ii) 2.7 kJ
Answer: (c) (i) 189 K (ii) 2.7 kJ
Solution: For an adiabatic process, PV^γ=constant (nRT/V)V^γ=constant or TV^(γ-1)=constant T1 V1^(γ-1)=T2 V2^(γ-1) T2=T1[V1 / V2]^(γ-1) Here, T1=27°C=300 K, V1=V, V2=2 V, γ=5 / 3 T2=300(V / 2 V)(5 / 3-1)=300(1 / 2)^(2 / 3)=189 K Change in internal energy, ΔU=nCv ΔT ΔU=n[(f / 2)R](T2−T1)=2 ×(3 / 2) ×(25 / 3)(189-300) ΔU=-2.7 kJ
Topic: Thermal Expansion
Q6: The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 1000°C is (For steel Young's modulus is 2 × 10^11 Nm−2 and coefficient of thermal expansion is 1.1 × 10−5 K−1)
Options: (a) 2.2 × 10^7 Pa
(b) 2.2 × 10^6 Pa
(c) 2.2 × 10^8 Pa
(d) 2.2 × 10^9 Pa
Answer: (c) 2.2 × 10^8 Pa
Solution: Given, ΔT=100°C, Y=2 × 10^11 N m−2 α=1.1 × 10−5 K−1 Young's modulus, Y= stress/strain =(F / A) /ΔL / L Therefore, Y=P /ΔL / L Y=P / α ΔT [ since ΔL=L α ΔT] Thermal stress in a rod is the pressure due to the thermal strain. Required pressure P=Y α ΔT=2 × 10^11 × 1.1 × 10−5 × 100=2.2 × 10^8 Pa
Topic: Thermal Expansion
Q7: Two wires are made of the same material and have the same volume. However, wire 1 has cross-sectional area A and wire 2 has cross-sectional area 3A. If the length of wire 1 increases by Δx on applying force F, how much force is needed to stretch wire 2 by the same amount?
Options: (a) 6 F
(b) 9 F
(c) F
(d) 4 F
Answer: (d) 9 F
Solution: For the same material, Young's modulus is the same and it is given that the volume is the same and the area of the cross-section for the wire 11 is A and that of 12 is 3 A. V=V1=V2 V=A × 11=3 A × 12 12=11 / 3 Y=(F / A) /Δ1 / l ⇒F1=YA(Δ11 / 11) F2=Y 3 A(Δl2 / l2) Given Δl1=Δl2=Δx (for the same extension) F2=Y 3 A(Δx / 11 / 3)=9(YA Δx / 11)=9 F1 or 9 F
Topic: Thermal Expansion
Q8: A uniform cylindrical rod of length L and radius r, is made from a material whose Young's modulus of elasticity equals Y. When this rod is heated by temperature T and simultaneously subjected to a net longitudinal compressional force F, its length remains unchanged. The coefficient of volume expansion, of the material of the rod, is (nearly) equal to
Options: (a) 9 F / (π r^2 YT)
(b) F / (3 π r^2 YT)
(c) 6 F / (π r^2 YT)
(d) 3 F / (π r^2 YT)
Answer: (d) 3 F / (π r^2 YT)
Solution: Given, Length =L Longitudinal force =F Radius =r Young's modulus =Y Temperature =T Since length remains same (Stress)Compressive =( Stress ) Thermal F / A=Y α T ⇒F / π r^2=Y α T (where α= coeficient of linear expansion) α=F / π r 2 YT Coefficient of volume expansion, γ=3 α γ=3 F / π r^2 YT
Topic: Thermal Expansion
Q9: During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio CP / CV for the gas is
Options: (a) 4 / 3
(b) 2
(c) 5 / 3
(d) 3 / 2
Answer: (d) 3 / 2
Solution: In an adiabatic process, T^γ= (constant) P^γ-1 T^γ / γ-1= (constant)P Given T^3= (constant) P γ / γ-1=3 ⇒ 3 γ-3=γ ⇒ γ=3 / 2
Topic: Thermal Expansion
Q10: An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by
Options: (a) 3 / P α K
(b) P / 3 α K
(c) 3 α / PK
(d) 3 PK α
Answer: (b) P / 3 α K
Solution: The bulk modulus of the gas is given by K=-P /ΔV / V 0 (Here negative sign indicates the decrease in volume with pressure) ΔV / V0=P / K [in magnitude] Also, V=V0(1+γ ΔT) or ΔV / V0=γ ΔT Comparing eq. (1) and (2), we get P / K=γ ΔT ΔT=P / 3 α K (since γ=3 α )
Topic: Thermal Expansion
Q11: Steel wire of length ' L ' at 40°C is suspended from the ceiling and then a mass ' m ' is hung from its free end. The wire is cooled down from 40°C to 30°C to regain its original length ' L '. The coefficient of linear thermal expansion of the steel is 10−5 /°C, Young's modulus of steel is 10^11 N / m^2 and radius of the wire is 1 mm. Assume that L > diameter of the wire. Then the value of ' m ' in kg is nearly
Answer: 3
Solution: E=(F / A) /ΔL / L =(F × L)/(A × ΔL) ⇒(ΔL / L)=F / AE Since, (ΔL / L)=α Δt α Δt=F / AE α Δt=mg / AE m=[(α Δt) / g] AE m=3.14 ≈ 3
Topic: Thermal Expansion
Q12: A diatomic gas with rigid molecules does 10 J of work when expanded at constant pressure. What would be the heat energy absorbed by the gas, in this process?
Options: (a) 25 J
(b) 35 J
(c) 30 J
(d) 40 J
Answer: (b) 35 J
Solution: The value of Cp for a diatomic gas is Cp=(7 / 2) R-(1) It is given that work done during expansion at constant pressure is dW=10 J=nR ΔT Since the gas undergoes an isobaric process ⇒ ΔQ=n(7 / 2) R ΔT=7 / 2(nR ΔT)=(7 / 2)(10)=35 J