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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A ring of radius R rolls without sliding with a constant velocity. The radius of curvature of the path followed by any particle of the ring at-the highest point its path will be : |
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Answer» R |
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| 2. |
The velocity vec(v) of a particle of mass 'm' acted upon by a constant force is given by vec(v)(t)=A[cos(kt)bar(i)-sin(kt)bar(j)]. Then the angle between the force and the velocity of the particle is (Here A and k are constants) |
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Answer» `90^(@)` |
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| 3. |
Give an example of a case where the velocity of an object is zero but its accleration is not zero? |
| Answer» Solution :In case OFA BODY PROJECTED VERTICALLY up. | |
| 4. |
Find the average of magnitude of linear momentum of helium molecules in a sample of helium gas at temperature of 150 piK. Mass of a helium molecules =(166//3) xx 10^(-27) kg and R = 25//3J-mol^(-1)K^(-1) |
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Answer» |
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| 5. |
A particle of mass m describes uniform circular motion in a horizontal plane. The quantitythat is conserved is |
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Answer» LINEAR velocity |
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| 6. |
A car can move at a maximum speed of 72 kmcdot h^(-1) on a curved road of radius of curvature 100 m. what is the coefficient of friction between the tyres of the car and the road? |
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| 7. |
If the vectors vec(P)= a hat(i) + a hat(j) + 3hat(k) and vec(Q)= a hat(i) - 2hat(j) - hat(k) are perpendicular to each other then the positive value of 'a' is |
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Answer» zero |
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| 8. |
Suppose, the acceleration due to gravity at the Earth.s surface is 10 m//s^2 and at the surface of Mars it is 4.0 m//s^2. A 60 kg passenger goes from the earth to the mars in a spaceship moving with a constant velocity. Neglect all other objects in the sky. Which part of figure best represents the weight (net gravitational force) of the passenger as a function of time. |
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Answer» A |
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| 9. |
A body of mass 1 kg is suspended from a weightless spring having force constant 600N*m^(-1). Another body of 0.5kg moving vertically upwards hits the suspended body with a velocity of 3m*s^(-1) and gets embedded in it. Find the frequency of oscillations and amplitude of motion. |
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| 10. |
The equivalance of two systems in thermal equilibrium is represented by the property of |
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Answer» heat |
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| 11. |
Why do rail tracks have a 'I shaped girder ? |
| Answer» SOLUTION :For REDUCING the EFFECT of BUCKLING, they are KEPT as .l-shaped.. | |
| 12. |
A lift which an acceleration 12 m//s^(2) downwards has an inclined plane rough at an angle 30^(@) as shown. A uniform ring of mass m and radius R, is placed at rest w.r.t lift at the middle of the roof. The coefficient of friction between ring and roof is sqrt(3)//5. The ring w.r.t lift will (takeg =10" m/s"^(2)) |
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Answer» roll with SLIPPING TOWARDS A |
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| 13. |
The radius of a well is 7 m. Water in it is at a depth of 20 m and depth of water column is 10m. Work done in pumping out water completely from the well is, (g=10 ms^(-2)) |
| Answer» ANSWER :D | |
| 14. |
Figure given below shows the displace show ment versus time graph for two particles A and B executing simple harmonic motions. Find the ratio of their maximum velocities, |
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Answer» Solution :For A, time period `T_(A)` = 16s (Distance between two adjacent crets) for B, time period `T_(B) = (26-2) = 24s` [LENGTH between the CREST and trough SHOWN = 20s-8s 12s] Also, amplitudes `a_(A) = 10cm, a_(B) = 5cm` Ratio of maximum velocities `(V_(A))/(V_(B))= (a_(A)omega_(A))/(a_(B)omega_(B))= (a_(A)T_(B))/(a_(B)T_(A))= (10 xx 24)/(5 xx 16)= (3)/(1)` |
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| 15. |
Centripetal force is …… to move a body ….. In a….. |
| Answer» SOLUTION :the FORCE REQUIRED , UNIFORMLY , CIRCLE . | |
| 16. |
Two bodies A and B of masses 10 kg and 5 kg are placed very slightly separated as shown in figure . The coefficient of friction between the floor and the blocks are mu_(s) = mu_(k) = 0.4 . Block A is pushed by an external force F . The value of F can be changed . When the welding between block A and ground breaks , block A will start pressing block B and when welding of B also breaks , block B will start pressing the vertical wall . If F = 20 N ,with how much force does block A presses the block B ? |
Answer» Solution :As `F LT f_(L) (mu N)` 
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| 17. |
A bullet fired into a fixed target loses half of its velocity in penetrating 15 cm. How much further it will penetrate before coming to rest ? |
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Answer» 5 cm |
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| 18. |
Two bodies A and B of masses 10 kg and 5 kg are placed very slightly separated as shown in figure . The coefficient of friction between the floor and the blocks are mu_(s) = mu_(k) = 0.4 . Block A is pushed by an external force F . The value of F can be changed . When the welding between block A and ground breaks , block A will start pressing block B and when welding of B also breaks , block B will start pressing the vertical wall . If F = 50 N , the friction force acting between block B and ground will be : |
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Answer» 10 N as `"" F gt f_(L)` So CONTACT FORCE between A and B is 10 N . |
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| 20. |
Two bodies A and B of masses 10 kg and 5 kg are placed very slightly separated as shown in figure . The coefficient of friction between the floor and the blocks are mu_(s) = mu_(k) = 0.4 . Block A is pushed by an external force F . The value of F can be changed . When the welding between block A and ground breaks , block A will start pressing block B and when welding of B also breaks , block B will start pressing the vertical wall . The force of friction acting on B varies with the applied force F according to curve : |
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Answer»
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| 21. |
For two bodies A and B of same material held on a horizontal plane force of limitting friction F varsus normal reaction R graphs are as shown in Which one has smoother surface in contact with the plane . . |
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Answer» `A` `:.` Surface of body B in contact with the plane must be smoother |
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| 22. |
A body of mass m is executing SHM with amplitude a. When its displacement x=1 unit, the force is b then what will be its maximum kinetic energy? |
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Answer» Solution :Restoring force at distance x= 1, `F= KX = k = B` `therefore b = k` Now MAXIMUM kinetic BODY of oscillator body `=(1)/(2)kA^(2) = (1)/(2) bA^(2)`. |
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| 23. |
A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this spring, when displaced and released, oscillates with period of 0.60 s. What is the weight of the body? |
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Answer» Solution :Here, m =50kg, Max. extension,y=20-0=20 cm=0.2 m,T=0.6s, Max. Force. F = mg =`50xx9.8 N therefore K = F/y = (50xx9.8)/(0.2) =2450 Nm^(-1)` Therefore As `T= 2pi SQRT(m/k) , m = (T^2 k)/(4pi^2) = ((0.6)^2 xx 2450)/(4XX(3.14)^2) = 22.36`kg `therefore ` Weight of BODY = mg `= 22.36 xx 9.8 = 219.1 N` |
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| 24. |
(A) : The distance of a star from earth can be measured by parallax method. (R) : The change in position of an object due to change in the point of observation is called parallax. |
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Answer» Both (A) and (R) are TRUE and (R) is the correct explanation of (A) |
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| 25. |
The area under the force, displacement curve is |
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Answer» POTENTIAL energy |
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| 26. |
A pendulum is secured on a cart rolling without friction down an inclined plane of inclination alpha . The period of the pendulum on an immobile cart is T_0How will the period of the pendulum change when the cart rolls down are slope ? |
| Answer» SOLUTION :`(T_0)/(SQRT(COS ALPHA))` | |
| 27. |
Calculate the moments of inertia of the figures shown, each having mass M, radius R and having uniform mass distribution about an axis perpendicular to the plane and passing through the centre? |
| Answer» Solution :`(MR^(2))/2`. These CASES are same as CIRCULAR disc. The distributios with RESPECT to axis of rotation is same PATTERN as circular disc. | |
| 28. |
A solid whose volume does not change with temperature, floats in a liquid. Fractions f_(1)andf_(2) of its volume remain submerged for two different temperatures t_(1)andt_(2) of liquid. The volume coefficient of expansion of liquid is |
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Answer» `( f_1 + f_2)/( f_1 t_1 + f_2 t_2)` |
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| 29. |
The diameter of a hose pipe is 2 cm. 1m^(3) of water discharged from the pipe in 20 s and it hits a wall. If the entire kinetic energy of water is converted into heat and this heat is absorbed by the water, what will be the increase in temperature of the water? J = 4.2 J cdot cal^(-1). |
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| 30. |
A uniform bar 8m long has a mass of 6kg. A 10kg mass is tied to one end. The position of centre of mass of the system is |
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Answer» `1.5m` from 10kg MASS |
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| 31. |
Oxygen and hydrogen gases are at the same temperature the ratio of the average K.E of an oxygen molecule and that of a hydrogen molecule is |
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Answer» 16 |
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| 32. |
A particle of mass 1 g moves on a straight line. The variation of its velocity with time is shown in fig. Find the force acting on the particle at t = 1 s, 4 s and 7 s. |
Answer» Solution : (i) At t = 1 s, acceleration of the particle is equal to the slope of the STRAIGHT line from 0 to 3s as 1 s is between 0 and 3s. Acceleration of the particle at t = 1 s is `a=(v_(f)-v_(i))/(t_(f)-t_(i))=(15-0)/(3-0)=5ms^(-2)` `therefore` Force ACTING on the particle `= ma =1xx10^(-3)xx5=5xx10^(-3)N`. (ii) At t = 4 s, acceleration is given by a = slope of the straight line from t = 3 to 5 s `= (v_(f)-v_(i))/(t_(f)-t_(i))=(15-15)/(5-3)=(0)/(2)=0 ms^(-2)` `therefore` Force acting on the particle `= ma = 1xx10^(-3)xx10=0N` (iii)At t = 7 s, acceleration is given by a = slope of the straight line from t = 5 to 8s `= (v_(f)-v_(i))/(t_(f)-t_(i))=(0-15)/(8-5)=(15)/(3)=-5 ms^(-2)` `therefore` Force acting on the particle `= ma = 1xx10^(-3)xx-5=-5xx10^(-3)N`. Negative sign indicates that the force is in the OPPOSITE direction of the motion of the particle. |
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| 33. |
What is a heat engine? Explain its working with schematic diagram. |
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Answer» Solution :Heat engine is a thermodynamics DEVICE which converts heat energy into mechanical energy (work). Working of Cornot's engine: Step 1: The working substance (ideal gas) is enclosed in a non-conducting wall and conducting bottom of a cyclinder fitted with air tight non conducting piston. This is placed on the source having in infinite thermal capacity at a steady temperature. The top surface is conducting and the rest non conducting. As a result, the gas expands isothermally. The work done by the system, `W_1= mu RT_1 log.(V_2)/(V_1) ("Curve A B")` Step 2: The working substance is now placed on a non conducting platform, as a result of which no heat exchange takes place between the system and the surroundings. The system expands adiabatically at the expense of its internal energy. The gas COOLS. The work done by the system, `W_2(mu R)/(gamma -1)(T_1-T_2) ("Curve B C")`  Step 3: The working substance is now placed on the sink maintained at a steady low temperature `T_2 K.` The system undergoes isothermal compression at this temperature. The pressure of gas increases and volume decreases without any change in internal energy and specific heat of gas REMAIN at infinity. `W_3=mu RT_2 log((V_4)/(v_3)) ("Curve C A")` Step 4 : The working substance is placed on a non-conducting platform. Under thermal isolation, the system undergoes change in its internal energy and its specific heat REMAINS at zero. Adiable compression results in increase in the pressure and temperature at the expense of work being done on the system. The system is allowed to reach its initial state. This completes one cycle of operation. The area bounded by the curves gives the amount of heat converted into work. `W_4=(-mu R)/(gamma-1)(T_1-T_2) ("Curve D A")`. i.e., Net work done `= W= W_1+W_2+W_3+W_4`. Work done `=mu R(T-T_2) log_e((V_2)/(V_1))`. 
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| 34. |
Name three physical quantities without dimensions. |
| Answer» SOLUTION :1. Strain 2. ANGLE 3. RELATIVE density | |
| 35. |
a point source of light is rotating in a horizontal plane at a speed of omega radians/second. There is a wall at a distance d from the source. At some instant the focus of the light is at P and angleSPN=theta (see figure). Speed of the focus at this instant in terms of theta is |
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Answer» `omegad//costheta` |
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| 36. |
A lift is moving vertically upwards with an acceleration a. What will be the changed time period of a simple pendulum suspended from the roof of the lift? If the lift becomes free and starts falling down with the acceleration due to gravity, what will be the change in the time period? |
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Answer» SOLUTION :ACCELERATION due to gravity (downwards) = g Acceleration of the LIFT (upwards) = -a Hence, the effective acceleration due to gravity with respect to the lift, `g.=g-(-a)=g+a` Therefore, the time period of the pendulum in the lift, `T.=2pisqrt(L/(g+a))` We compare it with the time period inside the lift with no acceleration: `T=2pisqrt(L/g)` `therefore" "T/(T.)=sqrt((g+a)/g)or,T.=Tsqrt(g/(g+a))` As `glt(g+a),T.ltT`, hence, in this CASE, the time period will decrease. When the lift is falling freely downwards, its acceleration a = g. Hence, the acceleration due to gravity with respect to the lift, g. = g - a = g - g = 0 Hence, the time period, in this case, `T.=2pisqrt(L/(g.))` = infinity. Infinite time period signifies that the pendulum will not oscillate at all. |
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| 37. |
Distinguish between the terms gravitation and gravity. |
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Answer» Solution :GRAVITATION: It is the force of attraction between any two BODIES in the universe. GRAVITY: It is the force of attraction between the EARTH and any object LYING on or near its surface. |
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| 38. |
A number of small drops of mercury coalesce adiabatically to form a single drop. The temeprature of drop |
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Answer» DECREASES |
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| 39. |
The fulcrum of a lever of length 1 m is at its middle point. If the fulcrum is moved by 0.25 m towards the loads , how are efficiency and velocity ratio affected? |
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| 40. |
The inner circumference of a steel tyre is 157 cm. It is to be fitted on a ring of diameter 50 cm. Find the temperature up to which the tyre is to be heated to fit it on the ring. When cooled, what force will be exerted by the tyre on unit surface area of the ring?For steel Y=2.1 times 10^(12)dyn*cm^(-2), alpha=12 times 10^(-6@)C^(-1). |
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| 41. |
Below are four equations expressing relationships between physical qualities. .g.gravitational field strengths, r radius, rhodensity, G the universal constant of gravitation, m a mass and T a time period. Which of the equations are true. a) gT^(2) = 4pi r^2 b) (4)/3 pirhoGr^(2) = g c) gr^(2) = Gm d) g=(4)/3pirhoGr |
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Answer» only a&B are TRUE |
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| 42. |
From a sphere of radius 1m, a sphere of radius 0.5 m is removed from the edge. The shift in the C.M. is |
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Answer» 13/16 m |
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| 43. |
A copper calorimeter of mass 600 gm contains 500 gm of water at temperature of 20^(@)C. A 500 gm of copper block at 100^(@)C is dropped into calorimeter. If the resultant temperature is 25^(0)C, find the specific heat of copper. |
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Answer» SOLUTION :Mass of copper calorimeter `(m_(1)) = 300 gm = 0.3` Kg , Specific heat copper = `S_(1)` Change in temp of the calorimeter `= 25 - 20 = 5^(@)C ,` Heat gained by the calorimeter `= m_(1) S_(1)(5) = 0.3 xx S_(1) xx 5 = 1.55_(1)J` Mass of water `(m_(2)) = 500 gm = 0.5` Kg , Specific heat of water `(S_(2)) = 4186 J Kg^(-1)K^(-1)` Change in temp of the water `= 25 – 20 = 5^(@)C` , Heat gained by the water `= m_(2) S_(2)5 = 0.5 xx 4186 xx 5 = 10465J` Mass of the copper block `(m_(3))`= 500 gm = 0.5 Kg, Specific heat of copper `= S_(1)` Change in temp of the block `= 100 – 25 = 75^(@)C` Heat lost by the copper block `m_(3)S_(1) xx 75 = 0.5 xx S_(1) xx 75= 37.5S_(1)J` According to the method of mixtures , Heat lost by copper = Heat gained by the calorimeter + Heat gained by the water `37.5 S_(1) = 1.5 S_(1) + 10465` `369S_(1) = 10465 or S_(1) = (10465)/(36) = 290.7` `therefore" Specific heat of copper "= 290.7 J Kg^(-1)K^(-1.)` |
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| 44. |
A ball reaches a rocket at 60 m/s along + X direction, and leaves the rocket in the opposite direction with the same speed. Assuming that the mass of the ball as 50 gm and the contact time is 0.02 second, the force exerted by the racket on the ball is |
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Answer» 300 N ALONG + X DIRECTION |
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| 45. |
A thin metal disc of mass 0.5 kg and cross sectional area 0.05 m^2is placed coaxially with a cylindrical block of height 0.5 m and same cross sectional area. The top face of the cylindrical block is maintained at a constant temperature of 300 K. If the initial temperature of the metal disc is 200 K, how long will it take for its temperature to raise to 250 K? The thermal conductivity of cylinder's material is 10 Wm^(-1) K^(-1)and specific heat capacity of disc's metal is 600 J kg^(-1) K^(-1)Note: Consider that the thermal conductivity of disc is very high and the system is thermally insulated except the top face of cylinder. |
Answer» SOLUTION :The given situation is shown below:  Here T is the temperature of the disc at any time t. The rate of heat flowing to the disc is `(dQ)/(dt) = KA (T_0 - T) // h `...(i) Due to flow of this heat, temperature of the disc increases by dT. Then, `(dQ)/(dt) =ms ( (dT)/(dt) )`...(ii) Comparing (i) and (ii), we get `KA (T_0 - T) // h = ms ((dT)/(dt) )` ` rArr (dT)/(dt)=(KA)/(hms) (T_0 -T)` Integrating above equation, we get ` int_(T_1)^(T_2) (dT)/(T_0 - T) = (KA)/(hms_0) int_(0)^(t) dt` where `T_1` and `T_2`are the initial and final temperatures of the metal disc, RESPECTIVELY, and t is the time taken ` rArr t = (hms)/(KA) ln ( (T_0 - T_1)/(T_0 - T_2) )` Here , `T_1 = 200 K` `T_2 = 250 K, T_0 = 300 K ` h = 0.5 m, m = 0.5 KG `s = 600 J kg^(-1) K^(-1)` `K = 10 W m^(-1) K^(-1)` ` A = 0.05 m^2` ` t = (0.5 xx 0.5 xx 600)/(10 xx 0.05)ln ( (300 - 200 )/(300 - 250) )` ` = 300 ln(2)= 208 s ` |
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| 46. |
A railway carriage of mass 9000 kg moving with a speed of 36kmh^(-1) collides with a stationary carriage of same mass. After the collision, the carriages get coupled and move together. What is their common speed after collision? What type of collision is this? |
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Answer» Solution :`m_(1)=9000kgkg,u_(1)=(36km)/H=(10 m)/s` `m_(2)=9000KG, u_(2)=0,v=v_(1)=v_(2)= ?` BY conservation of momentum : `m_(1)u__(1)+m_(2)u_(2)=(m_(1)+m_(2))v` `therefore v=(5m/s)` `"total K.E. before collision"=1/2m_(1)u_(2)^(2)+1/2m_(2)u_(2)^(2)=45000J` `"Total K.E. after collision =1/2(m_(1)+m_(2))v_(2)=225000J` `"As total K.E. after collision" lt "total K.E.before collision"` `therefore "collision is INELASTIC".` |
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| 47. |
A body is projected horizontally from the top of a tower. When the body strikes the ground its velocity is V an the direction of motion makes 30^(@) with the horizontal. Find the velocity of projection. |
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| 48. |
The distance travelled by a particle in time t is given by s=(2.5)t^(2).find (A)the average speed of the particle during the time 0 to 5.0s, and (B) the instantaneous speed at t=5.0s Here s is in metres. |
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Answer» Solution :The distance TRAVELLED during tiem `0"to"5.0s` is `=(2.5)(5.0)^(2)=62.5m.` the average SPEED during this TIME is `v_(av)=(62.5m)/(5s)=12.5m//s` `s=(2.5)t^(2)"or"(ds)/(dt)=(2.5)(2t)=(5.0)t` At `t=5.0s "the speed is" v=(ds)/(dt)=(5.0)(5.0)=25m//s` |
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| 49. |
According to kinetic theory of gases, absolute temperature of a gas is directly proportional to the average kinetic energy of translation per molecules to the gas. Zeroth law of thermodynamics established that temperature is the obly physical quantity that determines whether a given system is in thermal equilibrium with another system. heat flows from a system at higher temperatures to another system at lower temperature till their temperatures becomes equal. temperature corresponds to level in case of liquids, pressure in case of gases potential in case of electricity. Read the above passage and answer the following questions : (i) The absolute temperature of a gas is increased three times. what will be the increase in rms velocity of gas molecules ? What values of life do you learn this concept of temperature ? |
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Answer» Solution :(i) As `T prop 1/2 mC^(2) :. T prop C^(2) or C prop sqrt(T)` When T becomes three times , C becomes `sqrt(3)` times. `:. Increases in rms velocity = `sqrt(3) C - C = (1.732-1) C = 0.732 C = 73.2 %` (ii) Temperature is the degree of hotness of the system. just as a liquid flows higher to lower level, gas flows higher pressure to lower pressure, charge flows from higher to lower POTENTIAL , in the same way, HEAT energy flows from higher temperature to lower temperature. thus directions of flow of heat energy is determined by higher level of heating of the system and not by higher heat content of the system. This is so true in every sphere of day life. for example, an educational INSTITUTION is KNOWN not by the NUMBER of teaching faculty, but by the level of teaching faculty and the level of infrastructure provided. |
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| 50. |
Middle C on a piano has a fundamental of 262 Hz, and the first A above middle C has a fundamental frequency of 440 Hz. a. Calculate the frequencies of the next two harmonics of the C string . b. If A and C, strings have the same linear mass densitymu and length L, determine the ratio opf tensions in the two strings . |
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Answer» Solution :Remember that the harmonics of a vibrating string have frequencies that are related by integer multiples of the FUNDAMENTAL . This first part of the example is a sample SUBSTITUTION problem. Knowing that the frequency is ` f_(1) = 262 Hz` , find the frequencies of the next harmonics by multiplying by integers : ` f_(2) = 2f_(1) = 524 Hz` `f_(3) = 3f_(1) = 786 Hz` b. This part of the example is more of an analysis problem than is part (a). Use Eq. (III) to write expression for the fundamental frequencies of the two string. ` f_(1 A) = (1)/( 2 L) sqrt((T_(A))/( mu)) and f_(1) C = (1)/(2L) sqrt (( T_(c))/(mu))` Divide the first equation by the second and solve for the ratio of tensions `( f_(1)A)/( f_(1) C) = sqrt((T_(A))/(T_(C)))` `(T_(A))/( T_(C )) = ((f_(1) A)/( f_(1) C))^(2) = (( 440)/( 262))^(2) = 2.82` If the frequencies of piano strings were determined solely by tension . This result suggests that the ratio of tensions from the LOWEST string to the highest string on the piano would be enormous . Such large tensions would make it difficult to design a frame to support the strings . In reality , the frequencies of piano strings vary due to additional parameters , including the mass per unit length and the length of the string. |
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