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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 cylclist is riding with a speed of 27 km h^-1. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.5 ms^-2. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ? |
Answer» Solution :This is a case of a body moving in a circular path with variable speed. So it has centripetal acceleration `a_( c)`and tangential `a_(T)` which are perpendicular to each other. The RESULTANT acceleration.  `a = SQRT(a_( c)^(2) + a_(T)^(2)), a_( c) = v^(2)/r` `v = 27 km//h = (27 xx 1000)/(60 xx 60) = 7.5 ms^(-1), r= 80 m` `a_( c) = (7.5)^(2)/80= 0.703 ms^(-2)` Tangential acceleration `a_(T) = (Deltav)/(Deltat) = 0.5/1 = 0.5 ms^(-2)` `a = sqrt(a_(c )^(2) + a_(T)^(2)) = sqrt(0.703^(2) + 0.5^(2)) = 0.8627 ms^(-2)` Let the resultant acceleration make an angle a with the radius. Then `tan alpha = a_(T)/a_( c) = 0.5/(0.703) = 0.7112, alpha = tan^(-1) (0.7112) = 35^(@)30.`. |
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| 2. |
Which of the following parirs of physical quantities have same dimension? |
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Answer» TORQUE and Power |
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| 3. |
A block is projected with velocity v_(0) up the inclined plane from its bottom at t=0. The plane makes an angle theta with the horizontal. If the coefficient of friction between the block and the incline is mu = tan alpha(alpha gt theta) then fricitonal force applied by the plane on the block for t gt (v_(0))/(g[sin theta+tan alpha cos theta]) will be : |
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Answer» `TAN ALPHA mg COS THETA` |
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| 4. |
A golfer standing on the ground hits a ball with a velocity of 52m/s at an angle thetaabove the horizontal if tantheta=(5)/(12) find the time for which the ball is at least 15m above the ground? (g=10m//s^(2)) |
Answer» Solution :`v_(x) = SQRT(u_(y)^(2)-2gy)`  `= sqrt(52xx52xx(5xx5)/(13xx13) - 2xx10xx15) = sqrt(16xx25-300) = 10` `DELTAT = (2u_(y))/(10)= (2XX10)/(10) =` 2s |
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| 5. |
When a compass needle is held arbitrarily in any direction in earth's magnetic field, a couple acts on the needle. The force on north pole of needle is towards the north and force on south pole of the needle is towards south. These forces are equal in magnitude and opposite in direction. They from a couple, which rotates the compass needle and aligns it along north south direction. Once the alignment is complete, net force and net torque on the compass needle reduce to zero. Read the above passage and answer the following : (i) What is the origin of force and torque on the compass needle ? (ii) What values of life do you learn from this study ? |
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Answer» Solution :(i) Horizontal componentof earth's magnetic field is responsible for the force as well as torque on the compass needle. (ii) Once the compass needle alings itself ALONG the horizontal component of earth's magnetic field, the DEFLECTING forces and torque cases to act on the compass needle. In day to day life, when we GET aligned as per the wishes of four superiors (i.g., parents, teachers, bosses at work ), NOTHING is going to disturb US. |
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| 6. |
A particle of mass m moving along the x-axis is a potential energy U(x) = a + bx^2 where a and b are positive constants. It will execute simple harmonic motion with a frequency dete by the value of: |
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Answer» B ALONE |
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| 7. |
A rope of negligible mass is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rops is pulled with a force of 30 N? What is the linear acceleration of the rope? |
| Answer» SOLUTION :`ALPHA` = 25 `rads^(-2)`. a =10`ms^(-2)` | |
| 8. |
Using mass (M). Length (L), time (T) and electric current (A) as fundamental quantities the dimensions of permittivity will be |
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Answer» `MLT^(-1)A^(-1)` |
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| 9. |
A particle executes SHM with period T about a point O. It passes through a point P with velocity V along OP. Show that the time that elapses when it again comes to P is given by t = (T/pi)tan^(-)(TV)/(2pi.OP) |
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Answer» |
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| 10. |
Give two examples of the motion of big objects where the object can be treated as a particle and where it cannot be? |
| Answer» SOLUTION :In kinematics objects are TREATED as PARTICLES and their mass is not CONSIDERED and in dynamics the mass of the objects are considered | |
| 11. |
If the difference of two unit vectors is also a vector of unit magnitude, the magnitude of the sum of the two unit vectors is |
| Answer» Answer :C | |
| 12. |
The centripetal force required for a 1000 kg car travelling at 36 kmph to take a turn by 90^(@) in travelling along an arc of length 628 m is |
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Answer» 250 N |
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| 13. |
A rocket in space is being propelled by the ejection of a gas. Will the velocity of the centre of mass of the rocket be zero? |
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Answer» |
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| 14. |
A light body and a heavy body have the same kinetic energy. Which one will have greater momentum? |
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Answer» SOLUTION :GIVEN DATA `E_1 - E_2` `1/2 m_1v_1^2= 1/2 m_2v_2^2` or` (v_2^2)/(v_1^2) = m_1/m_2 " or" v_2/v_1 = sqrt(m_1/m_2)` As `P_2 = m_2v_2" and " P_1 = m_1v_1` `P_2/P_1 = (m_2v_2)/(m_1v_1) = m_2/m_1 sqrt(m_1/m_2) = sqrt( (m_2^2 m_1)/(m_1^2 m_2))` ` = P_2/P_1 = sqrt(m_2/m_1) , ` if ` m_2 GT m_1` then `P_2 gt P_1` i.e. a heavier body has GREATEST linear momentum |
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| 15. |
Explain sea breeze and ground waves. |
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Answer» Solution :During the day, the ground heats up more quickly than large bodies of water do. This occurs both because the water has a greater specific heat and mixing currents disperse the ABSORBED heat throughout the great volume of water. The air in contact with the warm ground is heated by conduction. It expands. It expands, becoming less dense than the surrounding COOLER air. As a result, the warm air rises (air currents) and other air moves (winds) to FILL the space - creating a sea breeze NEAR a large body of water. Cooler air descends, and a thermal convection cycle is set up, which TRANSFERS heat away from the land. At night, the ground loses its heat more quickly, and the water surface is warmer than the land. Sea breeze and ground waves as shown in figure.  
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| 16. |
A wire whose cross-sectional area is 2 mm^2 is stretched by 0.1 mm by a certain load. If a similar wire of triple. the area of cross-section is stretched by the same load, the elongation of the second wire would be |
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Answer» 0.33 MM |
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| 17. |
state the factors upon which the efficiency of a heat engine depends . |
| Answer» SOLUTION :it DEPENDS only on the TEMPERATURE of the SOURCE and the SINK | |
| 18. |
{:(,"Column I",,"Column II"),(("A"),GM_(e)M_(s),(p),["M"^(2)"L"^(2)"T"^(-3)]),(("B"),(3RT)/(M),(q),["ML"^(3)"T"^(-2)]),(("C"),F^(2)/q^(2)B^(2),(r),["L"^(2)"T"^(-2)]),(("D"),GM_(e)/R_(e),(s),"None"):} |
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| 19. |
Three identical masses are kept at the corners of an equilateral triangle ABC. A moves towards B with a velocity V, B moves towards C with velocity V, and C moves towards A with same velocity V. Then the velocity of centre of mass of the system of particles is |
| Answer» ANSWER :B | |
| 20. |
Two particles each of mass m are attached at end points of a massless rod AB of length l. Rod is hinged at point C as shown. Rod is released from rest from horizontal position. At the instant when rod reaches its vertical position as shown, which of the following is / are correct: |
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Answer» Speed of the particle at B is thrice the speed of particle at A |
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| 21. |
Two radio stations broadcast their programmes at the same amplitude A, and at slightly different frequenciesomega_(1) and omega_(2) respectively , whereomega_(2) - omega_(1) = 10^(3) Hz. A detector receives the signals from the two stations simultaneously . It can emit signals only of intensity ge 2 A^(2). (i). Find the time intervals between successive maxima of the intensity of the signal received by the detector . ii. Find the time for which the detector remains idle in each cycle of the intensity of the signal . |
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Answer» Solution :i. Let the signal waves be given by ` y_(1) A sin 2 pi omega_(1) t,y_(2) = A sin 2pi omega_(2) t`. The RESULTANT disturbance is given by ` y = y_(1) + y_(2) = A sin 2 pi omega_(1) t + A sin 2 pi omega_(2) t` `= 2 A sin ( 2 pi ( omega_(1) + omega_(2)) t)/( 2) cos ( 2 pi ( omega_(2) - omega_(1))t)/(2)` `= 2 A cos pi ( omega_(2) - omega_(1)) t sin 2 pi (( omega_(1) + omega_(2)) t)/(2)` Let `omega_(1) = omega , omega_(2) = omega + Delta omega` Therefore , ` omega_(1) + omega_(2) ~~ 2 omega` ` y = 2 A cos pi ( omega _(2) - omega_(1))t sin 2 pi omega t` Thus , the resultant disturbance has amlitude ` 2A cos ( omega_(2) - omega_(1)) t` For maxima : `cos pi ( omega_(2) - omega_(1))t = +- 1` or ` pi ( omega_(2) - omega_(1)) t = r pi , r = 0 , 1,2 , 3` ` t = (r)/( omega_(2) - omega_(1)) = 0 , (1)/( omega_(2) - omega_(1)) , (2) /( omega_(2) - omega_(1) ,...` CLEARLY time interval between successive maxima ` = (1)/( omega_(2) - omega_(1)) = (1)/( 10^(3)) = 10^(-3) s` ii. The resultant intensity is given by `I = A_(1) ^(2) + A_(2)^(2) + 2 A_(1) A_(2) cos delta` when ` delta = 0 ` , intensity is maximum `I_(max) = 4 A^(2)` When `Delta = pi//2, "intensity" I = 2 A^(2)` When ` Delta = pi , "intensity" I_(min) = 0` When `Delta = 3pi//2, "intensity" I = 2A^(2)` When `Delta = 2pi , "intensity" I_(max) = 4 A^(2)` The DETECTOR remaining idlefrom ` theta = pi//2 to 3 pi //2` or in each half cycle . Hence the required time ` t = (T)/(2) = (10^(-3))/(2) = 5 xx 10^(-4) s` |
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| 22. |
Distinguish between streamline and thrbulent flow of liquids. |
| Answer» Solution : When a liquid flows such that each PARTICLE of the liquid passing a point moves ALONG the same path and has same velocity as the preceding particle, its FLOW is called stream line flow. When a liquid flows such taht the motion of any particle of the liquid at any point varies rapidly in magnitude and DIRECTION, its flows is called turblent flow Turblent flow is known as unsteady flow. In this type of flow the velocity of the liquid at any GIVEN point varies with time. | |
| 23. |
Breaking stress of a wire depends on |
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Answer» the RADIUS of the wire |
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| 24. |
The planet is observed from two diametrically opposite points A and B on Eath. The angle theta subtended at the planet by the two directions of observation is 1^(@)8'. Given the diameter of the Earth to be about 1.276xx10^(7)m, computer the distanc |
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Answer» |
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| 25. |
There are two holes one each along the opposite sides of a wide rectangular tank. The cross-section of each hole is 0.01 m^2 and the vertical distance between the holes is 1 m. The tank is filled with water. The net force on the tank in newton when the water flows out of the holes is (Density of water = 1000 kgm^(-3)) |
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Answer» 100 |
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| 26. |
The planet is observed from two diametrically opposite points A and B on Eath. The angle theta subtended at the planet by the two directions of observation is 1^(@)30'. Given the diameter of the Earth to be about 1.276xx10^(7)m, computer the distance of the moon from the Earth. |
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Answer» |
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| 27. |
Consider the x-axis as representing east, the y-axis as north and z-axis as vertically upwards. Give the vector representing each of the following points .5 m north east and 2 m up, |
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Answer» 5 m north east and 2 m up Length along Y-axis `=5sin45^(@)=(5)/(sqrt2)m` Length along Z-Axis =2m `:.` In vector rotation `=(5)/(sqrt2)hati+(5)/(sqrt2)hatj+2hatk=(5(hati+hatj))/(sqrt2)+2hatk` (b) Length along `X =4cos45^(@)=(4)/(sqrt2)m` Length along `Y=4sin45^(@)=(4)/(sqrt2)m` Length along Z-axis =3m In vector rotation `=(4)/(sqrt2)hati-(4)/(sqrt2)hatj+3hatk=4(hati-hatj)//sqrt2+3hatk` (c) Length along `X=-2cos45^(@)=-(-2)/(sqrt2)m=-sqrt2m` Length along `Y=2sin45^(@)=(2)/(sqrt2)m=sqrt2m` length along Z=4m `:.` In vector rotation `=-sqrt2hati+sqrt2hatj+4hatk=(-i+j)sqrt2+4hatk`. 
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| 28. |
When two pieces of ice are pressed together, they combine to form one single piece. This process is called. |
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Answer» Freezing |
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| 29. |
The rate of change of total momentum of a system of many particle system is proportional to the cdots on the system |
| Answer» Answer :A | |
| 30. |
A rocket of mass 20 kg has 180 kg of fuel. The exhaust velocity of fuel is 1.6 km/sec. Calculate the ultimate velocity of the rocket gained, when the rate of consumption of the fuel is 2kg/sec. (neglect gravity) |
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Answer» 3.7 km/sec |
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| 31. |
The unit 1 N m^(-) is equivalent to |
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Answer» `1 erg cm^(-1)` `:.`Equivalent unit is `Jm^(-2)` |
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| 32. |
In a heat engine, the temperature of the source and sink are are 500K and 375 K. If the engine consumes 25xx10^(5)J per cycle, the work done per cycle is |
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Answer» Solution :`T_(1) = 500 K, T_(2) = 375K,` Heat absorbed `= Q_(1) = 25 xx 10 ^(5) J` `ETA = (T_(1) -T_(2))/(T_(1)) = (500 - 375)/(500) = -0.25 = 25%` ` eta = (W)/(Q_(1))` `W = eta xx Q_(1) = 0.25 xx 25 xx 10 ^(5) J` `= 6. 25 xx 10 ^(5) J ` per cycle Heat REJECTED to the sink per cycle `=Q_(2) = ?` `Q_(2) = Q_(1) - W = 25 xx 10 ^(5) -6.25 xx 10 ^(5)` `= 18.75 xx 10 ^(5) J` |
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| 33. |
A uniform cylinder of height h and radius r is placed with its circular face on a rough inclined plane and the inclination of the plane to the horizontal gradually in creased. If mu is the coefficeint of friction, then under what conditions the cylinder a) slide before toppingb) topple before sliding |
Answer» Solution : a) The CYLINDER will SLIDE if `MgsinthetagtmuMgcosthetaimpliestanthetagtmu.......(1)` The cylinder will topple if `(Mgsintheta)(h)/(2)gt(Mgcostheta)rimpliestanthetagt(2r)/(h).........(2)` THUS, the condition of sliding is `tanthetagtmu` and condition of TOPPLING `tanthetagt(2r)/(h)` Hence, the cylinder will slide before toppling if `MULT(2r)/(h)` b) The cylinder will topple before sliding if `mugt(2r)/(h)` |
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| 34. |
At constant pressure, the ratio of increase in volume of an ideal gas per degree rise in Kelvin temperature to its original volume is |
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Answer» `sqrt(3)` |
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| 35. |
A solid disc and a ring, both of radius 10cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10pi rad s^(-1). Which of the two will starts to roll earlier ? The co-efficient of kinetic friction is mu_(k)=0.2. |
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Answer» Solution :Frictional force acting on the body, `f=mu_(k)R=mu_(k)mg`………..`(i)` where `R(=mg)` is the NORMAL reaction of the table on the body. If a is the acceleration of the CM of the body `f=ma`…………`(II)` From eqns. (i) and (ii) `ma=mu_(k)mg` or `a=mu_(k)g`............`(iii)` Torque due to frictional force i.e., `tau=fr=(mu_(k)mg)r=mu_(k)mgr` As `tau=Ialpha.alpha=(tau)/(I)=(mu_(k)mgr)/(1)` Linear velocity of CM of the body (initially at rest, `v_(0)=0`) is given by `v=v_(0)+at=at=muk"gt"`.............`(iv)` Angular velocity of the body after time t, i.e. `omega=omega_(0)+alphat=omega_(0)-((mu_(k)mgr)/(I))t`...........(v) (Here `alpha` has been taken NEGATIVE as torque due to frictional force produces retardation) `v=romega`...........(vi) From eqs. (iv), (v) and (vi) we get `mu_(k)"gt"=r[omega_(0)-(mu_(k)mgr)/(I)]t=romega_(0)-((mu_(k)mgr^(2))/(I))t`................(vii) For a ring , as `I=mr^(2)`, from eqn. (vii) `mu_(k)"gt"=romega_(0)-((mu_(k)mgr^(2))/(mr^(2)))t=romega_(0)-mu_(k)"gt"` or `2mu_(k)"gt"=romega_(0)` or `t=(romega_(0))/(2mu_(k)g)` or `t=(0.1xx10pi)/(2xx0.2xx9.8)=0.80s` (as `r=0.1omega_(0)=10pirad//s`, `mu_(k)=0.2`, `g=9.8m//s^(2)`) or `3mu_(k)"gt"=romega_(0)` or `t=(romega_(0))/(3mu_(k)g)` or `t=(0.1xx10pi)/(3xx0.2xx9.8)=0.53s` Since t is less in case of disc the disc begin to roll earlier than the ring for the same VALUES of r and `omega_(0)` |
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| 36. |
A thin wire ring of 1m. Is situated on the surface of a liquid. If the excess force required to lift it upwards (before the liquid film breaks) from the liquid surface is 8N, then the surface tension of liquid is :-(in N/m) |
| Answer» ANSWER :B | |
| 37. |
A copper ball (specific heat = 0.1 CGS units) weighing 100 gm is heated in furnace until it attains temperature of the fumace. It is then dropped in to 2Kg oil (specific heat = 0.8 CGS units) at 35^(0)C. The temperature of the oil raises by 5^(0)C. Then find temperature of the furnace. |
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Answer» |
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| 38. |
The pulleys A and C are fixed while the pulley B is movable . A mass M_(2) is attached to pulley B , while the string has masses M_(1) "and " M_(3) at the two ends . Find the acceleration of each mass . |
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Answer» Solution :The force acting on masses `M_(1),M_(2)"and " M_(3)` are SHOWN in figure . Let `a_(1) "and" a_(3)` be upward acceleration of masses `M_(1)"and" M_(3) "and" a_(2)` the downward acceleration of mass `M_(2)`. For motion of mass M_(1) `T-M_(1) G=M_(1)a_(1)`....(1) For motion of mass `M_(3)` `T-M_(3)g=M_(3) a_(3)`....(2) For motion of mass `M_(2)` `M_(2)g-2T=M_(2)a_(2)`....(3) Let `l_(1),l_(2),l_(3)` be the lengths of vertical portions of the string between PULLEY at any instant . `M_(1) ` goes up through `x_(2) M_(2)` goes down through y and `M_(3)` goes up through z. Then `l_(1)+2l_(2)+l_(3)=(l_(1)-x)+2(l_(2)+y)+l_(3)-z` `rArr x+z =2y rArr (d^(2)x)/(dt^(2))+(d^(2)z)/(dt^(2))=2(d^(2)y)/(dt^(2))` `a_(1)+a_(3)=2a_(2) rArr a_(2)=(a_(1)+a_(3))/2`....(4) From (1), a_(1)=T/M_(3)-g`....(5) From (2),` a_(3)=T/M_(3)-g`....(6) substituting `a_(1) "and" a_(3)` in (4) , we get `a_(2)=T/2(1/M_(1)+1/M_(3))-g`....(7) From (3), `T=(m_(2)(g-a_(2))/2`....(8) substituting this value in (7), we get `a_(2)=M_(2)/4 (g-a_(2))(1/M_(1)+1/M_(3))-g "or" a_(2)=(M_(2)(g-a_(2))(M_(3)+M_(1)))/(4M_(1)M_(3))-g` or `4M_(1)M_(3)a_(2)-M_(2)(M_(1)+M_(3))g-M_(2)(M_(1)+M_(3)a_(2)-4M_(1)M_(3)g` or `[4M_(1)M_(3)+M_(2)(M_(1)+M_(3))]a_(2)=(M_(1)+M_(3))M_(2)g-4M_(1)M_(3)g` `a_(2)=((M_(1)+M_(3))M_(2)g-4M_(1)M_(3)g)/(4M_(1)M_(3)+M_(2)(M_(1)+M_(3)))a_(2)=(M_(1)M_(2)+M_(2)M_(3)-4M_(1)M_(3))/(M_(1)M_(2)+M_(2)M_(3)+4M_(1)M_(3))....(9) Substituting value of `a_(2)` in above equation , we get `a_(1)=(-M_(1)M_(2)+3M_(2)M_(3)-4M_(1)M_(3))/(M_(1)M_(2)+M_(2)M_(3)+4M_(1)M_(3))g`....(10) similarly from (6)and (9) `a_(3)=(3M_(1)M_(2)-M_(2)M_(3)-4M_(1)M_(3))/(M_(1)M_(2)+M_(2)M_(3)-4M_(1)M_(3))`....(11) 
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| 39. |
Find the external workdone on the system in Kcal, when 12.5 k cal of heat is supplied to the system and the corresponding increasing in internal energy is 10500 J (J = 4200 J/kcal) |
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Answer» 15 K CAL |
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| 40. |
What is the difference between physical and biological sciences ? |
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Answer» SOLUTION :Physical sciences deal with the proparties and behaviour of non LIVING matter. BIOLOGICAL sciences deal with living THINGS. |
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| 41. |
A particle moves along a circular path under the action of a force. The work done by the force is |
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Answer» POSITIVE and non-ZERO |
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| 42. |
The work done by the conservative force in a cycle is |
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Answer» zero |
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| 43. |
If x is the displacement of a simple harmonic oscillatorat a certain instant and y its acceleration at that instant, draw graph of (x, y) at all instant in one complete oscillation, |
Answer» SOLUTION :The acceleration y of a particle executing shm at.any in stant is `y = - omega^x, x` is the displacement at that instant and wo is a constant. Thus y is proportional to x and `omega^2`hence the GRAPH is a straight line passing through the origin as SHOWN in the Fig. For the second half of the oscillation the CURVE is RETRACED. When `x = -a, y =omega^2 a`and when `x= a,y = - omega^2` awhere .a. is the amplitude. 
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| 44. |
A man cycles up a hill whoes slope is 1 m in 25 at the rate of 10 km/hr.the weight of the man and the cycle is 100 kg.Find the power at which he is working. |
| Answer» SOLUTION :109 WATT | |
| 45. |
A non-uniform ball of spherically symmetric mass distribution, radius 'R' and radius of gyration about geometric centre = R/2, is kept on a frictionless. The geometric centre coincides with the centre of mass. The ball is struck horizontally with a sharp impulse = J The point of application of the impulse is at a height 'h' above the surface. Then : |
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Answer» the ball will slip on surface for all cases |
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| 46. |
A train moving at a speed of 220 m/s towards a stationary object , emits a sound of frequency 1000 Hz . Some of the sound reaching the object gets reflected back to the train as echo . The frequency of the echo as detected by the driver of the train is |
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Answer» 3000 HZ |
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| 47. |
The dimensions of sigmab^(4) are (where sigma = Stefan's constant and b = Wein's constant) |
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Answer» `[M^(0)L^(0)T^(0)]` and `("ENERGY")/("Area" xx "Time")= SIGMA T^(4)` or `sigma= ("Energy")/(("Area" xx "Time")T^(4))` or `sigmab^(4)= (("Energy")/("Area" xx "Time"))lambda_(m)^(4)` `:. [sigmab^(4)]= ([ML^(2)T^(-2)])/([L^(2)][T])[L^(4)]= [ML^(4)T^(-3)]` |
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| 48. |
A physics professor was driving a Maruti car which has its rear wind screen inclined attheta = 37^(@) to the horizontal. Suddenly it started raining with rain drops falling vertically. After some time the rain stopped and the professor found that the rear wind shield was absolutely dry. He knew that, during the period it was raining, his car was moving at a constant speed of V_(c) = 20 km//hr.[tan 37^(@) = 0.75] (a) The professor calculated the maximum speed of vertically falling raindrops as V_(max). What is value of Vmax that he obtained. (b) Plot the minimum driving speed of the car vs. angle of rear wind screen with horizontal (theta) so as to keep rain off the rear glass. Assume that rain drops fall at constant speed V_(r) |
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Answer» |
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| 49. |
Estiamtethe meanabsolute errorfrom thefollowingdate . 20.17,21.23,20.79,22.07,21.78 |
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Answer» 0.85 |
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| 50. |
Assertion : It is better to wash clothes by soap with cold liquid. Reason : The hot liquid of soap have less surface tension than cold one . |
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Answer» Both ASSERTION and REASON are TRUE and the reason is the correct explanation of the assertion. |
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