JEE Advanced Physics Mock Test – 1
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Jee Advanced Physics Mock Test -1
Total Questions: 20
Total Marks: 80
Duration: 48 Minutes
- Correct Answer : 4 Marks
- Wrong Answer: -1 Mark
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Question 1 of 20
1. Question
A body of mass $3 , \text{kg}$ is dropped from a height of $20 , \text{m}$. What is its kinetic energy just before hitting the ground? (Take $g = 10 , \text{m/s}^2$)
Correct
Kinetic energy $KE = mgh = (3)(10)(20) = 600 , \text{J}$.
Incorrect
Kinetic energy $KE = mgh = (3)(10)(20) = 600 , \text{J}$.
Unattempted
Kinetic energy $KE = mgh = (3)(10)(20) = 600 , \text{J}$.
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Question 2 of 20
2. Question
The moment of inertia of a hollow sphere of mass $5 , \text{kg}$ and radius $0.3 , \text{m}$ about its diameter is:
Correct
Moment of inertia $I = \frac{2}{3} m r^2 = \frac{2}{3}(5)(0.3)^2 = 0.3 , \text{kg.m}^2$.
Incorrect
Moment of inertia $I = \frac{2}{3} m r^2 = \frac{2}{3}(5)(0.3)^2 = 0.3 , \text{kg.m}^2$.
Unattempted
Moment of inertia $I = \frac{2}{3} m r^2 = \frac{2}{3}(5)(0.3)^2 = 0.3 , \text{kg.m}^2$.
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Question 3 of 20
3. Question
A Carnot engine operates between $800 , \text{K}$ and $400 , \text{K}$. What is its efficiency?
Correct
Efficiency $\eta = 1 – \frac{T_2}{T_1} = 1 – \frac{400}{800} = 0.5 = 50%$.
Incorrect
Efficiency $\eta = 1 – \frac{T_2}{T_1} = 1 – \frac{400}{800} = 0.5 = 50%$.
Unattempted
Efficiency $\eta = 1 – \frac{T_2}{T_1} = 1 – \frac{400}{800} = 0.5 = 50%$.
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Question 4 of 20
4. Question
A parallel plate capacitor has plates of area $2 , \text{m}^2$ separated by a distance of $0.01 , \text{m}$ in air. What is its capacitance?
Correct
Capacitance $C = \frac{\epsilon_0 A}{d} = \frac{(8.85 \times 10^{-12})(2)}{0.01} = 1.77 \times 10^{-9} , \text{F}$.
Incorrect
Capacitance $C = \frac{\epsilon_0 A}{d} = \frac{(8.85 \times 10^{-12})(2)}{0.01} = 1.77 \times 10^{-9} , \text{F}$.
Unattempted
Capacitance $C = \frac{\epsilon_0 A}{d} = \frac{(8.85 \times 10^{-12})(2)}{0.01} = 1.77 \times 10^{-9} , \text{F}$.
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Question 5 of 20
5. Question
A body of mass $4 , \text{kg}$ is moving with a velocity $\vec{v} = 2\hat{i} + 3\hat{j} , \text{m/s}$. What is its momentum?
Correct
Momentum $\vec{p} = m\vec{v} = 4(2\hat{i} + 3\hat{j}) = 8\hat{i} + 12\hat{j} , \text{kg.m/s}$.
Incorrect
Momentum $\vec{p} = m\vec{v} = 4(2\hat{i} + 3\hat{j}) = 8\hat{i} + 12\hat{j} , \text{kg.m/s}$.
Unattempted
Momentum $\vec{p} = m\vec{v} = 4(2\hat{i} + 3\hat{j}) = 8\hat{i} + 12\hat{j} , \text{kg.m/s}$.
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Question 6 of 20
6. Question
A rod of length $2 , \text{m}$ and mass $8 , \text{kg}$ is pivoted at one end. What is its torque due to gravity when the rod is horizontal?
Correct
Torque $\tau = m g \frac{L}{2} = (8)(10)(1) = 80 , \text{Nm}$.
Incorrect
Torque $\tau = m g \frac{L}{2} = (8)(10)(1) = 80 , \text{Nm}$.
Unattempted
Torque $\tau = m g \frac{L}{2} = (8)(10)(1) = 80 , \text{Nm}$.
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Question 7 of 20
7. Question
A current of $10 , \text{A}$ flows through a resistor of $20 , \Omega$. How much heat is generated in $5 , \text{minutes}$?
Correct
Heat $H = I^2 R t = (10)^2 (20)(300) = 600,000 , \text{J}$.
Incorrect
Heat $H = I^2 R t = (10)^2 (20)(300) = 600,000 , \text{J}$.
Unattempted
Heat $H = I^2 R t = (10)^2 (20)(300) = 600,000 , \text{J}$.
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Question 8 of 20
8. Question
A capacitor of $10 , \mu\text{F}$ is connected in series with a $20 , \mu\text{F}$ capacitor. What is the equivalent capacitance?
Correct
Equivalent capacitance $C_{\text{eq}} = \frac{1}{\frac{1}{C_1} + \frac{1}{C_2}} = \frac{1}{\frac{1}{10} + \frac{1}{20}} = 6.67 , \mu\text{F}$.
Incorrect
Equivalent capacitance $C_{\text{eq}} = \frac{1}{\frac{1}{C_1} + \frac{1}{C_2}} = \frac{1}{\frac{1}{10} + \frac{1}{20}} = 6.67 , \mu\text{F}$.
Unattempted
Equivalent capacitance $C_{\text{eq}} = \frac{1}{\frac{1}{C_1} + \frac{1}{C_2}} = \frac{1}{\frac{1}{10} + \frac{1}{20}} = 6.67 , \mu\text{F}$.
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Question 9 of 20
9. Question
A charged particle moves in a circular path of radius $0.5 , \text{m}$ in a magnetic field of $0.2 , \text{T}$. If its velocity is $2 , \text{m/s}$, what is the charge-to-mass ratio $\frac{q}{m}$?
Correct
Radius $r = \frac{mv}{qB} \implies \frac{q}{m} = \frac{v}{rB} = \frac{2}{(0.5)(0.2)} = 20 , \text{C/kg}$.
Incorrect
Radius $r = \frac{mv}{qB} \implies \frac{q}{m} = \frac{v}{rB} = \frac{2}{(0.5)(0.2)} = 20 , \text{C/kg}$.
Unattempted
Radius $r = \frac{mv}{qB} \implies \frac{q}{m} = \frac{v}{rB} = \frac{2}{(0.5)(0.2)} = 20 , \text{C/kg}$.
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Question 10 of 20
10. Question
A photon has a wavelength of $600 , \text{nm}$. What is its energy?
Correct
Energy $E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{600 \times 10^{-9}} \approx 3.31 \times 10^{-19} , \text{J}$.
Incorrect
Energy $E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{600 \times 10^{-9}} \approx 3.31 \times 10^{-19} , \text{J}$.
Unattempted
Energy $E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{600 \times 10^{-9}} \approx 3.31 \times 10^{-19} , \text{J}$.
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Question 11 of 20
11. Question
A Carnot engine absorbs $1500 , \text{J}$ of heat and rejects $600 , \text{J}$. What is its efficiency?
Correct
Efficiency $\eta = \frac{W}{Q_{\text{in}}} = \frac{Q_{\text{in}} – Q_{\text{out}}}{Q_{\text{in}}} = \frac{1500 – 600}{1500} = 0.6 = 60%$.
Incorrect
Efficiency $\eta = \frac{W}{Q_{\text{in}}} = \frac{Q_{\text{in}} – Q_{\text{out}}}{Q_{\text{in}}} = \frac{1500 – 600}{1500} = 0.6 = 60%$.
Unattempted
Efficiency $\eta = \frac{W}{Q_{\text{in}}} = \frac{Q_{\text{in}} – Q_{\text{out}}}{Q_{\text{in}}} = \frac{1500 – 600}{1500} = 0.6 = 60%$.
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Question 12 of 20
12. Question
A sound wave in air has a frequency of $440 , \text{Hz}$ and a speed of $343 , \text{m/s}$. What is its wavelength?
Correct
Wavelength $\lambda = \frac{v}{f} = \frac{343}{440} \approx 0.78 , \text{m}$.
Incorrect
Wavelength $\lambda = \frac{v}{f} = \frac{343}{440} \approx 0.78 , \text{m}$.
Unattempted
Wavelength $\lambda = \frac{v}{f} = \frac{343}{440} \approx 0.78 , \text{m}$.
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Question 13 of 20
13. Question
A particle of mass $0.5 , \text{kg}$ moves in a circle of radius $2 , \text{m}$ with an angular velocity of $4 , \text{rad/s}$. What is its kinetic energy?
Correct
Kinetic energy $KE = \frac{1}{2} m \omega^2 r^2 = \frac{1}{2}(0.5)(4)^2 (2)^2 = 16 , \text{J}$.
Incorrect
Kinetic energy $KE = \frac{1}{2} m \omega^2 r^2 = \frac{1}{2}(0.5)(4)^2 (2)^2 = 16 , \text{J}$.
Unattempted
Kinetic energy $KE = \frac{1}{2} m \omega^2 r^2 = \frac{1}{2}(0.5)(4)^2 (2)^2 = 16 , \text{J}$.
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Question 14 of 20
14. Question
A capacitor of $8 , \mu\text{F}$ is charged to $12 , \text{V}$. What is the energy stored in the capacitor?
Correct
Energy $U = \frac{1}{2} C V^2 = \frac{1}{2} (8 \times 10^{-6})(12)^2 = 0.576 , \text{mJ}$.
Incorrect
Energy $U = \frac{1}{2} C V^2 = \frac{1}{2} (8 \times 10^{-6})(12)^2 = 0.576 , \text{mJ}$.
Unattempted
Energy $U = \frac{1}{2} C V^2 = \frac{1}{2} (8 \times 10^{-6})(12)^2 = 0.576 , \text{mJ}$.
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Question 15 of 20
15. Question
A Carnot engine absorbs $2000 , \text{J}$ of heat at $500 , \text{K}$ and rejects $1500 , \text{J}$ at $300 , \text{K}$. What is its efficiency?
Correct
Efficiency $\eta = \frac{W}{Q_{\text{in}}} = \frac{Q_{\text{in}} – Q_{\text{out}}}{Q_{\text{in}}} = \frac{2000 – 1500}{2000} = 0.25 = 25%$.
Incorrect
Efficiency $\eta = \frac{W}{Q_{\text{in}}} = \frac{Q_{\text{in}} – Q_{\text{out}}}{Q_{\text{in}}} = \frac{2000 – 1500}{2000} = 0.25 = 25%$.
Unattempted
Efficiency $\eta = \frac{W}{Q_{\text{in}}} = \frac{Q_{\text{in}} – Q_{\text{out}}}{Q_{\text{in}}} = \frac{2000 – 1500}{2000} = 0.25 = 25%$.
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Question 16 of 20
16. Question
A current of $3 , \text{A}$ flows through a resistor of $5 , \Omega$ for $2 , \text{minutes}$. How much heat is generated?
Correct
Heat $H = I^2 R t = (3)^2 (5)(120) = 5400 , \text{J}$.
Incorrect
Heat $H = I^2 R t = (3)^2 (5)(120) = 5400 , \text{J}$.
Unattempted
Heat $H = I^2 R t = (3)^2 (5)(120) = 5400 , \text{J}$.
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Question 17 of 20
17. Question
A point charge $q = 10^{-6} , \text{C}$ is placed at a distance of $2 , \text{m}$ from another charge $q‘ = -2 \times 10^{-6} , \text{C}$. What is the force between them?
Correct
Force $F = \frac{1}{4\pi \epsilon_0} \frac{q q‘}{r^2} = \frac{(9 \times 10^9)(10^{-6})(2 \times 10^{-6})}{(2)^2} = 4.5 , \text{N}$.
Incorrect
Force $F = \frac{1}{4\pi \epsilon_0} \frac{q q‘}{r^2} = \frac{(9 \times 10^9)(10^{-6})(2 \times 10^{-6})}{(2)^2} = 4.5 , \text{N}$.
Unattempted
Force $F = \frac{1}{4\pi \epsilon_0} \frac{q q‘}{r^2} = \frac{(9 \times 10^9)(10^{-6})(2 \times 10^{-6})}{(2)^2} = 4.5 , \text{N}$.
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Question 18 of 20
18. Question
A photon has a frequency of $6 \times 10^{14} , \text{Hz}$. What is its energy?
Correct
Energy $E = hf = (6.63 \times 10^{-34})(6 \times 10^{14}) = 3.978 \times 10^{-19} , \text{J}$.
Incorrect
Energy $E = hf = (6.63 \times 10^{-34})(6 \times 10^{14}) = 3.978 \times 10^{-19} , \text{J}$.
Unattempted
Energy $E = hf = (6.63 \times 10^{-34})(6 \times 10^{14}) = 3.978 \times 10^{-19} , \text{J}$.
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Question 19 of 20
19. Question
A solenoid of length $0.5 , \text{m}$ has $1000$ turns and carries a current of $2 , \text{A}$. What is the magnetic field inside the solenoid?
Correct
Magnetic field $B = \mu_0 n I = (4\pi \times 10^{-7})\left(\frac{1000}{0.5}\right)(2) = 5.03 \times 10^{-3} , \text{T}$.
Incorrect
Magnetic field $B = \mu_0 n I = (4\pi \times 10^{-7})\left(\frac{1000}{0.5}\right)(2) = 5.03 \times 10^{-3} , \text{T}$.
Unattempted
Magnetic field $B = \mu_0 n I = (4\pi \times 10^{-7})\left(\frac{1000}{0.5}\right)(2) = 5.03 \times 10^{-3} , \text{T}$.
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Question 20 of 20
20. Question
A transformer increases the voltage from $110 , \text{V}$ to $220 , \text{V}$. If the primary coil has $50$ turns, how many turns are in the secondary coil?
Correct
Voltage ratio $\frac{V_s}{V_p} = \frac{N_s}{N_p} \implies N_s = \frac{V_s}{V_p} \cdot N_p = \frac{220}{110} \cdot 50 = 100$.
Incorrect
Voltage ratio $\frac{V_s}{V_p} = \frac{N_s}{N_p} \implies N_s = \frac{V_s}{V_p} \cdot N_p = \frac{220}{110} \cdot 50 = 100$.
Unattempted
Voltage ratio $\frac{V_s}{V_p} = \frac{N_s}{N_p} \implies N_s = \frac{V_s}{V_p} \cdot N_p = \frac{220}{110} \cdot 50 = 100$.