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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. |
Derive an expression for orbital velocity of the satellite. |
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Answer» Solution :For a satellite of mass M to move in a circular orbit, CENTRIPETAL FORCE must be ACTING on the satellite. This centripetal force is provided by the EARTH's gravitational force. `(Mv^2)/((R_(E)+h))=(GMM_E)/((R_(E)+h)^(2))` `V^(2)=(GM_E)/((R_(E)+h))` `v=sqrt((GM_E)/((R_(E)+h)))` |
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| 2. |
If three particle each of mass m are placed at the three corners of an equilational triangle of side I, the work done to increase the side of that triangle 2I is |
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Answer» `(3Gm^2)/(L)` |
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| 3. |
The efficiency of a carnot engine oprerations between boiling freezing points of water is …. |
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Answer» 0.1 `eta=0.268` `eta=0.27` |
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| 4. |
Two bodies are projected from the same points withequal speeds in such directions that they both strike the same point on a plane whose inclination is beta. If alpha be the anlge of projection of the first body with the horizontal show that the ratio of their times of light is (sin(alpha-beta))/(cos alpha) |
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Answer» SOLUTION :Let `alpha` be the ANGLE of projection of the second body `R=(u^(2))/(g cos^(2)BETA)[sin(2 alpha-beta)-sin beta]` Range of both the bodies is same, therefore `sin (2alpha-beta)=sin(2 alpha.-beta)` or `sin 2 alpha.-beta=pi-(2alpha-beta)`,  `alpha.=(pi)/2-(alpha-beta)` Now `T=(2U sin (alpha-beta))/(g cos beta)` and `T.=(2u sin (alpha.-beta))/(g cos beta)` `:.T/(T.)=(2SIN (alpha-beta))/(2sin (alpha.-beta))=(sin(alpha-beta))/(sin{(pi)/2-(alpha-beta)-beta})` `sin=((alpha-beta))/(sin((pi)/2-alpha))=(sin(alpha-beta))/(cos alpha)` |
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| 5. |
A stone is dropped from a height of 10 cm above the top of a window 80 cm high. The time taken by the stone to cross the window is (g=10ms^(-2)) |
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Answer» `1/7`s |
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| 6. |
Two projectiles P and Q are fired from the same point with same velocity at angle 30^(@)and 60^(@). The horizontal range of : |
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Answer» <P>P is equal to Q `R_(P) = (u^(2))/(g) sin 60^(@)` `R_(Q) = (u^(2))/(g) sin 120^(@)` `= (u^(2))/(g) sin (180 - 60)` `R_(Q) = (u^(2))/(g) sin 60^(@)` `:. |R_(P)| = |R_(Q)|` |
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| 7. |
The moment of inertia of a cylinder about its own axis is eqal to its moment of inertia about an axis passing through its centre and normal to its length. The ratio of length to radius is |
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Answer» `2:1` |
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| 8. |
A solid sphere is rolling without slipping on rough ground as shown in figure. It collides elastically with an identical another sphere at rest. There is no friction between two spheres. Radius of each sphere is R and mass is m. Then linear velocity of first sphere (A) after it start rolling without slipping is |
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Answer» `(2)/(3)omegaR` |
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| 9. |
In SHM at the equilibrium position i) amplitude is minimum ii)acceleration is zero ii)velocity is maximum iv) potential energy is maximum |
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Answer» all are true |
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| 10. |
What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 xx 10 ^(3)kgm ^(-3)?Compressibility of water 45.8 10 ^(-11) Pa ^(-1), [1 Pa =1 Nm ^(-2)] |
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Answer» Solution :Let density of water at a depth is `rho.` and density of water at surface is `rho.` VOLUME of water of mass M and V at the surface and V. at a depth, `therefore V = (M)/(rho) and V. =(M)/(rho.)` `therefore` Change in volume `Delta =V-V.` `= M [(1)/(rho) - (1)/(rho.)]` `therefore (Delta V )/(V) = M [ (1)/(rho )- (1)/(rho ) ] xx (rho )/(M) = (rho )/(rho )-(rho )/(rho.) =1- (rho )/(rho.)` `therefore `Volume strain `(Delta V)/(V) =1 - (rho )/(rho.)` compressibility `K = (1)/("BULK MODULUS")` `= (Delta V )/(PV) = (1)/(P ) [1- (rho )/(rho.)]` `45.8 xx 10 ^(-11) = (1)/(80 xx 1.013 xx 10 ^(5)) [ (1- 1.03 xx 10 ^(3))/( rho .)]` `45.8 xx 10 ^(-11) xx 80 xx 1.013 xx 10 ^(5) =1 - ( 1.03 xx 10 ^(3))/(rho .)` `therefore (1.03 xx 10 ^(3))/( rho .) =1 - 45.8 xx 10 ^(-11) xx 80xx 1.013 xx 10 ^(5)` `therefore rho . =1- 3.712 xx 10 ^(-3) =0.996288` `therefore rho . = (1.03 xx 10 ^(3))/( 0.996288)` `therefore rho . = 1034 KG m ^(-3)` `therefore rho . =1.034 xx 10 ^(3) kg m ^(-3)` |
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| 11. |
Two particles having masses m and 2m are travelling along X-axis on a smooth surface with velocities u_(1) and u_(2) collide. If their velocities after collision are v_(1) and v_(2) then the ratio of velocities of their centre of mass before and after impact is |
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Answer» `2:1` |
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| 12. |
If v_1 is the sound in a diatomic gas at 273^@C and v_2 is the r.m.s. speed of its molecules at 273K, then (v_(1))/v_(2)= |
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Answer» `SQRT((15)/(14))` |
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| 13. |
A woman pushes a trunk on a railway platform which has a rough surface . She applies a force of 100 N over a distance of 10 m . Thereafter ,she gets progressively tired and her applied force reduces linearly with distance to 50 N . The total distance through which the trunk has been moved is 20 m . Plot the force appliedby the woman and the frictional force , which is 50 N versus dispalcement Calculate the work done by the two forces over 20 m . |
| Answer» SOLUTION :`W_(F) =1625 J , W_(f) = - 500 J ` | |
| 14. |
The speed of a car was 50 km//hr for the first 900s, then 40 km/hr for the 50 km and then the car decelerated uniformly at 10 km//hr^(2) till it came to rest. The average speed of the car was : |
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Answer» 50 km/hr `= ((v_(1)t_(1) + v_(2) t_(2) + (v_(2)^(2))/(2a_(2)))/(t_(1) + t_(2) + (v_(2))/(a_(2))))` |
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| 15. |
The velocity of a 2.5 kg block sliding down an inclined plane ( mu = 0.2) is found to be 1.5 "m.s"^(-1). One second later it has a velocity of 5 "m.s"^(-1). What is the angle of the plane with respect to the horizontal? |
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Answer» SOLUTION :Downward resultant force on the BLOCK along the inclined plane = mg sin `theta = muN = mgsintheta = mu COS theta` = mg (sin `theta -mu cos theta`) Downward acceleration, a = g(sin`theta -MUCOS theta) =9.8 (sin theta - 0.2 cos theta ) "m.s"^(-1)` From the relation v = u+at, a = `(v-u)/(t) = (1-1.5)/(1) = 3.5 "ms."^(-2)` `:. 9.8 (sin theta -0.2 cos theta) ` = 3.5 or, sin `theta-0.2 cos theta = (3.5)/(9.8) = (5)/(14)` Solving this equation we get `theta = 32^(@)` It is the angle of the plane with respect to the horizontal. 
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| 16. |
Three rods P, Q and R having identical shape and size and hinged togetherat ends to form an equilateral triangle. Rods P and Q are made of same material having coefficient of linear thermal expansion alpha_(1) while that of material of rod R is alpha_(2). By how any kelvin should the system of rods be increase the angle opposite to rod R by Delta theta. |
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Answer» <P> Solution :When the system is heated,the rods EXPAND and the triangle does not REMAIN equilateral. Let lengths of rods P, Q and R be `l_(1)l_(2) and l_(3)` respectively. Then from cosine rule,`cos theta = (l_(1)^(2) +l_(2)^(2) -l_(3)^(2))/(2l_(1)l_(2)) (or) 2l_(1)^(2) +l_(2)^(2)-l_(3)^(2)""....(a)`  Differentiating equation(a), we get `2l_(1) cos theta dl_(2) +2l_(2) cos theta dl_(1)-2l_(1)l_(2) sin theta d theta` `=2l_(1)dl_(1) +2l_(2) dl_(2) -2l_(3) dl_(3) "" ....(b)` Let the temperature of the system be increased by `DELTAT`. Then `dl_(1)=l_(1)alpha_(1) DeltaT` `dl_(2)=l_(2)alpha_(1) DeltaT and dl_(3)=l_(3) alpha_(2) DeltaT` In case of equilateral triangle `l_(1)=l_(2)=l_(3)=L` (say) and `theta=60^(@)` and for small change in angle `d theta =Delta theta` From equation (b) and (c ), we get `2l^(2)cos (60^(@)) alpha_(1) DeltaT +2l^(2) cos(60^(@)) alpha_(1)DeltaT-2l^(2) sin(60^(@))Delta theta` `=2l^(2)alpha_(1)DeltaT +2l^(2)alpha_(1)DeltaT-2l^(2)alpha_(2)DeltaT` `(alpha_(1))/(2)DeltaT +(alpha_(1))/(2)DeltaT-(sqrt(3))/(2) Delta theta=alpha_(1)DeltaT+alpha_(1) DeltaT-alpha_(2)DeltaT` `alpha_(1) DeltaT -(sqrt(3))/(2) Delta theta =2alpha_(1)DeltaT-alpha_(2) DeltaT` `(alpha_(1)-alpha_(2))DeltaT=(sqrt(3))/(2)Delta theta rArrDeltaT=(sqrt(3)Delta theta)/(2(alpha_(2)-alpha_(1)))` |
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| 17. |
A machine gun of mass 10 kg, fires 10 gm bullets at the rate of two every second. Each bullet comes out with a velocity of 200 m/s. The velocity of recoil of the gun at the end of the fourth second, after firing starts is |
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Answer» 1.6 m/s |
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| 18. |
The coefficient of restitution for a perfectly elastic collision is |
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Answer» 1 |
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| 19. |
The oscillation of a body on a smooth horizontal surface is represented by the equation, X=Acos (omega t) where X= displacement at time t omega = frequency of oscillation. Which one of the following graph shows correctly the variation a with ? |
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Answer»
`:.` VELOCITY `v=(DX)/(dt)=(d)/(dt)(A cos omegat)` `=-A omega sin omega t` Acceleration, `=a(DV)/(d)/(dt)(-A omega sin omegat)` `=-A omega^(2) cos ^(2) omega t :. A prop cos omega ` Hence the variation of a with t is correctly shown by graph (d). |
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| 20. |
A fully loaded boeing aircraft has a mass of 3.3 times 10^5 Kg. Its total wing area is 500 m^2. It is in level flight with a speed of 960 km//h (a) Estimate the pressure difference between the lower and upper surfaces of the wings (b) Estimate the fractional increase in the speed of the air on the surface of the wing relative to the lower surface.[ The density of air is p=1.2 kg m^-3] |
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Answer» Solution :(a) The weight of the BOEING aircraft is balanced by the upward force due to the pressure difference `Delta P times A=3.3 times 10^5 Kg times 9.8` `Delta P=(3.3 times 10^5 Kg times 9.8 ms^-2)//500 m^2` `=6.5 times 10^3 Nm^-2` (b) We ignore the small height difference between the top and BOTTOM sides in Eq.(10.12) the pressure difference between them is then `Delta P=p/2 (v_2^2-v_1^2)` Where `v_2` is the SPEED of air over the upper surface and `v_1` is the speed under the bottom surface. `(v_2-v_1)=(2 Delta P)/(p(v_2+v_1))` Taking the AVERAGE speed `v_(av)=(v_2+v_1)//2=960 km//h=267 ms^-1` we have `(v_2-v_1)//v_(av)=(Delta P)/(p v_(av)^2)=0.08` The speed above the wing needs to be only 8% higher than that below. |
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| 21. |
An electron and proton are detected in a cosmic ray experiment, the first with kinetic energy 10 KeV and the second with 100 KeV. Which is faster, the electron or the proton? Obtain the ratio of their speeds (electron mass : 9.11 xx 10^(-31)kg : proton mass : 1.67xx10^(-27) kg : 1 e v = 1.6 xx 10^(-19) J) |
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Answer» Solution :Here`K_e= 10eVand K_(p )= 100KeV` `m_(e )= 9.11 xx 10^(-31)Kg and m_p= 1.67xx 10^(-27 ) Kg ` as ` K =1/2mv ^2 ` or`v= sqrt((2K )/( m))` hence ` (v_p )/( V_p) = sqrt((K_e )/(K_p) xx (m_p)/(m_c))= sqrt((10 k e )/( 100ke V) xx ( 1.67 xx 10^(-27 ) kg )/( 9.11 xx 10^(-31) kg ) )= 13.54` ` V_(e )= 1354v_P` |
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| 22. |
The moment of inertia of a thin square plate ABCD of uniform thickness about an axis passing through the centre O and perpendicular to plate is |
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Answer» `l_(1)+l_(2)` |
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| 23. |
A person 'P' in train moving due south with 80 kmph and his friend Q moving due west at 60 kmph Q finds that P is travelling with a velocity of |
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Answer» 100 kmph, `Tan^(-1)((4)/(3))S` or E |
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| 24. |
A Uniform Metallic Rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperating slightly |
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Answer» Its speed of rotation increases |
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| 25. |
From a sphere of radius 1 m, 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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| 26. |
A bubble of soap water is at rest on a table in the dining compartment of a train, if the acceleration of the train is g/4 in forward direction, the angle made by its surface with horizontal is |
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Answer» `"TAN"^(-1) ((1)/(2))` |
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| 27. |
A particle is projected up into air from a point with a speed of20 m//s at an. angle of projection 30^@. What is the maximum height reached by it? |
| Answer» SOLUTION :`H=(mu^2sin^2theta)/(2G)=((20)^2xx(SIN30^@)^2)/(2xx9.8)=5.1m` | |
| 28. |
Small bits of camphor dance about in clear water but not on one contaminated with oil. Why? |
| Answer» Solution :When bits of camphor are placed in CLEAR WATER they decrease the surface tension on dissolving. The water near each such bit exerts a pull CAUSING an ERRATIC movement of the bit. In case of oil contaminated water the surfade tension is less than the surface tension DUE to dissolved camphor and thus no such movement takes place. | |
| 29. |
Two force are equal and opposite in direction exerted on a bar (rod) as shown in figure. Hence plane PQ makes an angle theta with the cross-section 'a'. Find the tensile stress on PQ. |
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Answer» Solution :In `DELTA PRQ, (PR)/(PQ) = COS THETA, ` Area of PR =a `therefore (a)/(PQ) = cos theta` `therefore` Tensile STRESS `= (F.)/(PQ)=(F cos theta)/(a //cos theta) = (F cos ^(2) theta )/(a)` |
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| 30. |
A gas (bulk modules =B) IN enclosed in a cylinder of volume V fitted with a piston of cross-section A and mass M. The piston is slightly displaced and released.Derive an expression for its time period. |
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Answer» Solution :`B=-V(DeltaP)/(DeltaV)` where `DeltaP=(F)/(A) and DeltaV=Ax, X=` dispacement of piston `rArr B=(-V(F)/(A))/(Ax)=(-VF)/(A^(2)x) rArr K=(F)/(x)=(BA^(2))/(V), T= 2pi SRT((m)/(k))= 2pi SQRT((M)/((BA^(2))/(V)))= 2pi sqrt((MV)/(BA^(2)))` |
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| 31. |
Animals curl into a ball, when they feel very cold. Why? |
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Answer» Solution :(d) When the animals feed cold, they curl their bodies into the ball so as to decrease the surface area of their bodies. As total energy RADIATED by a body VARIES directly as the surface area of the body, the loss of heat due to radiation would be reduced. (e) Normal human body TEMPERATURE (T) = `98.6^(@)F` Convert Fahrenheit into Kelvin, `(F - 32)/(180) = (K - 273)/(100)` So, `T = 98.6^(@)F = 310K` From Wien.s displacement law Maximum wavelength `lamda_("max") = b/T = (2.898 xx 10^(-3))/(310)` `lamda("max") = 9348 xx 10^(-9) m` `lamda("max") = 9348 "nm" (at 98.6^(@)F)` During HIGH FEVER, human body temperature `T = 104^(@) F = 313K` Peak wavelength `lamda_("max") = b/T = (2.898 xx 10^(-3))/(313)` `lamda_("max") 9259 xx 10^(-9) m` `lamda_("max") = 9259` nm (at `104^(@)F`) (f) When animals feel cold, they curl their bodies into the ball so as to decrease the surface area of their bodies. As total energy radiated by a body varies directly as the surface area of the body, the loss of heat due to radiation would be reduced. |
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| 32. |
The percentage increase in earth.s angular velocity so that all bodies lying on the equator feel weight lessness in nearly |
| Answer» Answer :C | |
| 33. |
The ratio of time taken by two objects A and B of masses 1 kg and 3 kg to free fall from height 16 m and 25 m respectively is ...... |
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Answer» `12/5` `THEREFORE d = 1/2 g t ^(2)` `t = SQRT ((2d)/(g))` it is independent of mass `therefore t _(1) = sqrt ((2 xx 16)/(9.8)) and t _(2) = sqrt ((2 xx 25)/(9.8))` `therefore (t_(1))/(t_(2)) = sqrt (( 2 xx 16)/(2 xx 2 5)) = 4/5` |
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| 34. |
Are inertial and gravitational mass of a body different from one another ? |
| Answer» SOLUTION :No, they are EQUIVALENT. | |
| 35. |
Vectors subtraction is ___________ |
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Answer» non-commutative only |
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| 36. |
A vector vec(A) is alongthe positive x-axis and its vector product with another vector vec(B) is zero, then vector vec(B) could be |
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Answer» `HAT(i) + hat(J)` |
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| 37. |
All types of SHM are periodic in nature Derive the mathematical expression for kinetic and potential energies of a particle executing simple harmonic motion. |
| Answer» Solution :A is amplitude.Amplitude is the maximum DISPLACEMENT from the mean position `OMEGA` is the ANGULAR velocity.Angular velocity is defined as the rate of change of angular displacement. | |
| 38. |
The muscles of a normal eye are least strained when the eye is focussed on an object |
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Answer» FAR AWAY from the eye |
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| 39. |
A particle starts from the origin at t = 0 s with a velocity of 10.0 hat(j)m//s and moves in plane with a constant acceleration of (8hat(i) + 2hat(j))ms^(-2). The y-coordinate of the particle in 2sec is ______ |
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Answer» 24m |
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| 40. |
A force of 0.1176 N acts horizontally for 5 s on a 50 g block kept on a floor. What is the velocity attained by the block? Coefficient of sliding friction betweenthe block and the floor is 0.2. |
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Answer» |
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| 41. |
The moment of inertia of a disc of mass M and radius R about an axis which is tangential to the circumference of the disc and parallel to the diameter is |
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Answer» `5/4 MR^2` |
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| 42. |
Two monkeys A and B are holding on the two sides of a light string passing over a smooth pulley. Mass of the two monkeys are m_(A)= 8 kg and m_(B) = 10 kg respectively [g = 10 m//s^(2)] (a) Monkey A holds the string tightly and B goes down with an acceleration a_(r) = 2 m//s^(2) relative to the string. Find the weight that A feels of his own body. (b) What is the weight experienced by two monkeys if A holds the string tightly and B goes down with an acceleration a_(r) = 4 m//s^(2) relative to the string. |
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Answer» (B) `(640)/(N) ` for both |
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| 43. |
A bullet of mass 10 g moving horizontally with a velocity of 400 ms^(-1) strikes a wook block of mass 2 kg which is suspended by light inextensible string of length 5 m. As a result, the centre of gravity of the block found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be |
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Answer» `100 ms^(-1)` Initial speed of bullet, `u = 400 ms^(-1)`. Mass of block , M = 2 kg LENGTH of string , l = 5 m Speed of the block after collision = `v_1` Speed of the bullet on emerging from block , `v = ?`  USING energy conservation principle for the block, `(KE + PE)_("Reference") = (KE + PE)_h` `implies 1/2 Mv_1^2 = Mgh " or " v_1 = sqrt(2gh)` `v_1 = sqrt(2 xx 10 xx 0.1) = sqrt(2) ms^(-1)` Using momentum conservation principle for block and bullet system, `(M xx 0 + m u)_("Before collision") = (M xx v_1 xx MV)_("After collision")` `implies 0.01 xx 400 = 2sqrt(2) + 0.01 xx v` `implies v = (4 - 2sqrt(2))/(0.01) =117.15 ms^(-1) ~~ 120 ms^(-1)` |
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| 44. |
What will be the effect on the value of the acceleration due to gravity on the earth's surface , if the radius of the earth suddenly reduces to half its present value and the mass remains constant ? |
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Answer» Solution :LET the mass of the earth be M andits radius be R. Acceleration due to gravity on the earth.s surface, `g=(GM)/(R^2)` When the radius CHANGES to `R/2` and the mass remains `g^.=(GM)/(R/2)^2=(4GM)/(R^2)` `therefore g^. =4g` Thus, the acceleration due to gravity would be 4 TIMES the PRESENT value. |
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| 45. |
Assertion At the same temperature and pressure, equal volums of all gases gases contain equal numbe of molecules. Reason In 1 L at NTP total number of molecules are 6.02xx10^(23). |
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Answer» If both Assertion and Reason are correct and Reason is the correct explanation of Assertion. |
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| 46. |
A ball is projected from the ground with speed u at an angle alphawith horizontal. It collides with a wall at a distance .a. from the point of projection and returns to its original position. Find the coefficient of restitution between the ball and the wall. |
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Answer» SOLUTION :As we have discussed in the theory, the horizontal component of the velocity of BALL during the path OAB is `u cos alpha ` while in its return journey BCO it is `eu coa alpha.` FTHE time of fight T also remains unchanged. Hence, `T = t _(OAB)+ t _(BCO) or (2 u sin alpha )/( g) = (a)/( u cos alpha ) + (a)/( eu cos alpha )` (or) `(a)/( eu cos alpha ) = (2 u sin alpha )/(g ) (a)/( ucos alpha )` (or )` (a )/( eu cos alpha ) = (2u ^(2) sin alpha cos alpha - ag)/(gu cos alpha)` `therefore e = (ag)/( 2u ^(2) sin alpha cos alpha - ag ) or e = (1)/(( ( u ^(2) sin 2 alpha )/( ag )-1))` 
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| 47. |
A point mass is tied to one end of a cord whose other end passes through a vertical hollow tube, caught in one hand. The point mass is being rotated in a horizontal circle of radius 2m with speed of 4 m/s. The cord is then pulled down so that the radius of the circle reduces to 1m. The ratio of kinetic energies under the final and initial state is |
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Answer» |
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| 48. |
If coefficient of real expansion of a liquid is 1//5500. the temperature at which its density is 1% less than density at 0^@ C is |
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Answer» `55.5^(@) C` |
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| 49. |
A sphere ,a cubeand a thin circular plate,all made of the same meteriel and having the same mass are initially heated to a temperature of 3000°C.Which of these will cool fastest |
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Answer» sphere |
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| 50. |
A disc of radius 'r' is removed from the disc of radius 'R' then |
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Answer» The minimum shift in centre of mass is ZERO |
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