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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. |
The ratio of the accelerations due to gravity at the bottom of a deep mine and that on the surface of the earth is 978/980. find the depth of the mine if the density of the earth is uniform throughout and the radius of the earth is 630 km. |
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Answer» 12.86 km |
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| 2. |
What is the reason for Brownian motion? |
| Answer» Solution :According to kinetic theory, any particle suspended in a LIQUID or gas is continuously bombarded from all the directions so that the MEAN free path is almost negligible. This LEADS to the motion of the particles in a random and zig-zag MANNER. | |
| 3. |
Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature T_(0), while box B contains one mole of helium at temperature (7//3)T_(0). The boxes are then put into thermal contact with each other , and heat flows between them until the gases reach a common final temperature (Ignore the heat capacity of boxes). Then , the final temperature of gases, T_(f) , in terms of T_(0) is - |
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Answer» `T_f=5/2T_0` but `DeltaE_"int"=0` `(DeltaE_"int")_(N_2)=(MU C_V DELTAT)_(N_2)(C_V)_(N_2)=5/2R` `(DeltaE_"int")_(He)=(mu C_V DeltaT)_(He) (C_V)_(He)=3/2R` From equation (1), `0=1xx5/2(T_f-T_0)+1xx3/2xx(T_f-7/3T_0)` `therefore 0=5 T_f-5T_0+3T_f - 7T_0` `therefore 12T_0 =8T_f` `therefore T_f=3/2T_0` |
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| 4. |
Let g be the acceleration due to gravity at earth's surface and K be the rotational kinetic energy of the earth. Suppose the earth's radius decreases by 2 % keeping all other quantities same, then |
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Answer» g decreases by2% and K decreases by 4% . `g prop 1/R^2 and K prop 1/R^2 "" ( :. I =2/5 MR^2)` |
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| 5. |
A barometer with brass scale which is correct at 0^(@)C reads 75 cm on a day at 20^(@)C. If Y_(real) of mercury - 0.00018 //^(@)C and a of brass is 0.000018//^(@)C, calculate the correct reading at 0^(@)C. |
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Answer» Solution :Let `H _(0)` be the correct height at `0^(@)C,` H be the observed height at `t ^(@)C.` Let `alpha and gamma` be the coefficients of LINEAR EXPANSION and cubical expansion of BRASS and MERCURY. Let `p _(0) and p ` be the densities of mercury at `0^(@)C and t ^(@)C.` If we use correction for the expansion of the scale and change in density of mercury. We can write `H (1 + alpha t ) p g = H _(0) p _(0) g ` `H (1 + alpha t ) p = H _(0) p (1 + rt) implies H_(0) ~~ H ( 1 + alpha t) (1- gamma t ) = H {1- (gamma - alpha) t}` But `H = 75 cm, t = 20^(@)C, So, H _(0) = 74.758 cm` |
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| 6. |
Determine the maximum acceleration of the trainin which a box lying on itsfloor will remain stationary given that the coefficeient of static friction between the box and the train 's floor is 0.15 takeg=10ms^(-2) |
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Answer» Solution :As the accelerationof the BOS ix due to staticfriction so ma `=F le mu_(s) R = mu_(3) mg` `:.,a = mu_(s) g` or `a_(MAX)=mu_(s) g= 0 .15xx 10= 1.5 MS^(-2)` |
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| 7. |
Two identical uniform rods AB and CD, each of length L are jointed to form an inverted T-shaped frame as shown in figure. Locate the centre of mass of the frame. The centre of mass of a uniform rod is at the middle point of the rod. |
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Answer» Solution :Let the mass of each rod be m. Take the centre C of the rod AB as the origin and CD as the Y-axis. The rod AB has mass m and its centre of mass is at C. For the calculation of the centre of mass of the combined system, AB may be replaced by a POINT particle of mass m placed at the point C. Similarly the rod CD may be replaced by a point particle of mass m placed at the centre E of the rod CD. Thus, the frame is equivalent to a system of two PARTICLES of equal masses m each, placed at C and E. The centre of mass of this pair of particle will be at the middle point F of CE. The centre of mass of the frame is, therefore, on the rod CD at a distance L/4 from C. 
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| 8. |
A steel rod has a radius of 10mm and a length of 1.0m A 100KN force stretches its along its length.Calculate the elongation of the steel rod. |
| Answer» SOLUTION :R=`10xx10^-3m, L=1m, F=100xx10^3N, Y=2xx10^11Nm^-3, DELTAL=?Y=FL//pir^2deltaL, deltaL=FL//pir^2Y=100xx10^3xx1//3.14xx(10xx10^-3)^2xx10^11 =1.59xx10^-3m` | |
| 9. |
In the figure light is incident on the thin lens as shown, the radius of curvature for both the surface is R. Determine the focal length of the system. |
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Answer» Solution :For refraction at FIRST SURFACE, `mu_2/v_1-mu_1/(-infty)= (mu_2-mu_1)/(+R)….(i)` For refraction at `2^(nd)` surface `mu_3/v_2=mu_2/v_1=(mu_3+mu_2)/(+R)`…….(ii) Adding equations (i) and (ii) we get `mu_3/v_2=(mu_3-mu_1)/R or v_2=(mu_3R)/(mu_3-mu_1)` Therefore focal length of the GIVEN lens system is `(mu_3R)/(mu_3-mu_1)` 
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| 10. |
A body is moving in a circular orbit. It is just about to slide to the outer and mu mg = (mv^(2))/(l). In this expression, mu represents |
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Answer» Coefficient of STATIC FRICTION |
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| 11. |
A ball of mass 0.2 kh is droppped down vertically from a height of 1m above the ground. If it rebounds to a height of 0.64m, find the coefficient of restitution between the surface of the ball and ground. |
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Answer» 0.32 |
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| 12. |
A sphere of radius R is supported by a rope attached to the wall. The rope makes an angletheta = 45^(@) with respect to the wall. The point where the rope is attached to the wall is at a distance of (3R)/(2) from the point where the sphere touches thewall. Find the minimum coefficient of friction (mu)between the wall and the sphere for this equilibrium to be possible. |
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Answer» |
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| 13. |
If the position vector of the particle is given by vecr=3hatt^(2)hati+5thatj+4hatk, Find the (a) The velocity of the particle at t=3s (b) Speed of the particle at t= 3s (c) Acceleration of the particle at time t=3s |
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Answer» Solution :The velocity `VECV=(dvecr)/(dt)=(dx)/(dt)hatj+(DZ)/(dt)hatk` We obtain `vecv(t)=6thati+5hatj` The velocity has only two components `v_(x)=6t`, depending on time and `v_(y)=5` which is independent of time. The velocity at t=3s is `vecv(3)=18hati+5hatj` (b) The speed at t=3s is `v=sqrt(18^(2)+5^(2))=sqrt(349)=18.68ms^(-1)` (c) The ACCELERATION `veca` is, `veca=(d^(2)vecr)/(dt^(2))=6hati` The acceleration has only the x-component. NOTE that acceleration here is independent of t, which means `veca` is constant. EVEN at 1=3s it has same value `veca+6hati`. The velocity is non-uniform, but the acceleration is uniform (constant) in this case. |
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| 14. |
Find the ratio of the orbital speeds to two satellites of the earth if the satellites are at height 6400km and 19200km. (Radius of the earth = 6400km.) |
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Answer» Solution :Orbital speed , `V_0=SQRT((GM)/(R+h))` The ratio of the orbital SPEEDS of TWO satellites AROUND the earth is `V_(01)/V_(02)=sqrt((R+h_2)/(R+h_1))` Where `h_1` and `h_2` are the heights of the satellites In this PROBLEM`h_1`=6400 Km, `h_2`= 19200 km,R=6400 km. `V_(01)/V_(02)=sqrt((6400+19200)/(6400+6400))=sqrt(25600/12800)=sqrt2/1` |
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| 15. |
Which of the following is not expressed correctly with respect to units? |
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Answer» Specific HEAT `rarr J kg^(-1) K^(-1)` |
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| 16. |
A metal rod of Young's modulus2 xx 10^10 Nm^(-2)undergoes an elastic strain of 0.02% the energy per unit volume stored in the rod in joule/m^3 is |
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Answer» 400 |
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| 17. |
A block of weight W is suspended from the mid - point of a rope whose ends are at the same horizontal level. The force required to straight the rope is |
| Answer» ANSWER :D | |
| 18. |
Along a vertical oily part a monkey rises for the first 3 s and then slips down for the next 3 s. The velocity of the monkey is given as v(t) = 2t(3-t), when 0 lt t lt 3 and v (t) = -(t-3)(6-t), when 3 lt t lt 6 Find (i) the average speed of the monkey (ii) time at which its speed is maximum. |
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| 19. |
Taking that earth revolves round the sun ina circular orbit of radius 15xx10^10 m, with a time period 1 yr the time taken by another planet , which is at a distance of 540 xx 10^10 m, to revolve rounnd the sun in circular orbit once , will be |
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Answer» 216 YEARS |
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| 20. |
Starting from rest a car takes at least ‘t’ second to travel through a distance s on a flat concrete road. Find the minimum time that will be needed for it to climb through a distance ‘s’ on an inclined concrete road. Assume that the car starts from rest and inclination of road is theta = 5^(@) with horizontal. Coefficient of friction between tyres and the concrete road is mu = 1. |
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Answer» |
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| 21. |
If linear momentumof a body is increases by 2 % its kinetic energy increases by ……… |
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Answer» <P>0.02 ` :. (DELTAK)/K xx 100 = 2 (Deltap)/p xx100 +(Deltam)/m ` but m is constant ` = 2xx2 ` % = 4 % |
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| 22. |
A number of small drop of mercury coalesce adiabatically to form a singledrop. The temperature of the drop. |
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Answer» increases |
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| 23. |
Keeping the velocity of projection constant, the angle of projection is increased from 0^(0) to 90^(@). Then the maximum height of the projectile |
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Answer» goes on INCREASING up to `90^(@)` |
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| 24. |
Keeping the velocity of projection constant, the angle of projection is increased from 0^(0) to 90^(@). Then the horizontal range of the projectile |
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Answer» GOES on INCREASING up to `90^(@)` |
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| 25. |
A body projected vertically from the earth reaches height equal to earth's radius before returning to the earth, The power exerted by the gravitational force is greatest |
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Answer» at the insant just before the BODY HITS the earth |
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| 26. |
Three rings each of mass P and radius R are arranged as shown in Figure. Then answer the following Suppose the whole arrangement comprising the three rings is turned through 120^(@) clockwise or anticlockwise about a vertical axis passing through the centroid of the triangle formed by joining the centres 1, 2 and 3 then moment of inertia of the system about YY' is |
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Answer» `7/2"PR"^(2)` |
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| 27. |
Three rings each of mass P and radius R are arranged as shown in Figure. Then answer the following Suppose the ring 2 is exactly superposed over ring 1, then the moment of inertia of the system about YY' will be |
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Answer» `7/2"PR"^(2)` |
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| 28. |
A body collides heat on and elastically with another identical body at rest then after collision |
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Answer» FIRST body COMES to rest, then second body begins to move with the INITIAL velocity of first body |
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| 29. |
Figureshows the position-time graph of a particle of mass 4 kg.What is the (a) force on the particle for t < 0, t > 4 s, 0 < t < 4 s? (b) impulse at t = 0 and t = 4 s ? (Consider one-dimensional motion only). |
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Answer» Solution : (a) In all the three intervals, acceleration and, therefore, force are zero. (b) `3KG ms^(-1) " at" t =0 ,( C) - 3 KG ms^(-1) " at" t = 4S` |
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| 30. |
Heat produced by a current is obtained a relation H = I^(2)RT. If the errors in measuring these quantities current, resistance, time are 1% 2% and 1% respectively then total error in calculating the energy produced is........... |
| Answer» SOLUTION :`5%` | |
| 31. |
"The friction between two surfaces increases when the surfaces are made smoother and smoother " . Why ? |
| Answer» SOLUTION :Because the area of contact INCREASES and HENCE the MOLECULAR force of attraction also increases. | |
| 32. |
Referring to figure calculate the downward acceleration of mass m_1. Assume the surfaces are frictionless and pullyes are massless. |
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Answer» Solution :LET a be the acceleration of mass M and `a_1 and a_2`, the acceleration of `M_1 and M_2` relative to fixed pulley, P. Then from SHOWN in figure. `M_1 g - T_2 = M_1 a_2` ........... (1) `M_2 g -T_2 = M_2a_2` ......... (2) `and T_1 = Ma ` Also `T_1 = 2T_2` ........ (4) Acceleration of `M_1` relative to movable pulley Q is `(a_1- a)` . Acceleration of `M_2` relative to pulley `Q=(a_2 -a)`. The acceleration of `M_1 and M_2` relative to puelly Q are equal and opposite . ` :. a_1 - a = -(a_2-a)` ` or a =(a_1 + a_2)/( 2)` .......... (5) substracting (2) from (1) `(M_1 - M_2) g = M_1 a_1 - M_2 a_2` ......... (6) Adding (1) and (2) , we get `(M_1 + M_2) g - 2T_2 = M_1 a_1 + M_2a_2` From (3) and (4) , ` 2T_2 = T_1 =Ma ` USING `(5) , 2T_2 =(M(a_1 + a_2))/(2)`............ (7) substituting this value in (7) , we get `(M_1 + M_2) g -(M(a_1+ a_2))/(2) = M_1 a_1 + M_2 a_2` ` or 2(M_1 + M_2) g - Ma_1 - Ma_2 = 2M_1 a_1 + 2M_2a_2` ` or 2(M_1 + M_2)g = (2 M_1 + M)a + (2M_2 + M)a_2` .......... (8) Eliminating ` a_2 ` from (6) and (8) , we get ` a_1 = [(4M_1 M_2 + M(M_1 -M_2))/(4M_1M_2+ M(M_1 + M_2))]g` 
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| 33. |
What meant by one candela? And Which base quantity is measured by this unit? (or) Define one candela (S.I standard for Luminous intensity) |
| Answer» Solution :One candela is the luminous INTENSITY in a given DIRECTION, of a source that emits monochromatic radiation of FREQUENCY `5.4 xx 10^(14) Hz` and that has a radiant intensity of `(1)/(683)` watt/steradian in that direction. | |
| 34. |
We are familiar with Newton's lows of motion. State Newton's second law of motion. |
| Answer» SOLUTION :REFER SECTION 5.5 | |
| 35. |
Is the bulb of a thermoeter made of diathermic or adiabatic wall ? |
| Answer» SOLUTION :The TEMPERATURE above which molar heat capacity of a solid SUBSTANCE becomes CONSTANT. | |
| 36. |
Rolling friction is ____ than both static friction and kinetic friction. [ Fill in the blanks ] |
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Answer» |
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| 37. |
The lower end of a capillary tube of radius r is placed vertically in water. Then with the rise of water in the capillary, heat evolved is : |
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Answer» `+(pi^(2)r^(2)H^(2))/(j)dg` |
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| 38. |
(A): For the planets orbiting around the sun, angular speed, linear speed, K.E. changes with time, but angular momentum remains constant. (R): No torque is acting on the rotating planet. So its angular momentum is constant. |
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Answer» Both ‘A’ and .R. are TRUE and .R. is the correct EXPLANATION of ‘A’ |
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| 39. |
{:("List -I","List -II"),((a)"it is possible to change the temperature of a gas without supplying heat ",(e)"isothermal process"),((b)"change in internal energy is zero",(f) "adiabatic process"),((c)"work done is zero ",(g)"isochoric process"),((d)"work done by the gas is maximum for a given increment in volume",(h)"isobaric process"):} |
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Answer» a-f, B-h, c-g, d-e |
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| 40. |
A student says that he had applied a force F= - r/x on a particle and the particle moved in simple harmonic motion. He refuses to tell whether r is a constant or not. Assume that he was worked only with positive x and no other force acted on the particle |
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Answer» As X increases K increases |
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| 41. |
A cyclindrical tube of length 100 cm Contains some mercury inside. If the length of remaining air column remains constant at all temperatures, the height of mercury is neglect the expansion of cross section (gamma_(B) of Hg= 18 xx 10^(-5)//^(@) C, gamma_(g) of tube = 2.7 xx 10^(-5)//^(@) C ) |
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Answer» 10 cm |
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| 42. |
A force vec(F)= (5hat(i) - 3hat(j) + 2hat(k))N moves a particle from vec(r )_(1)= (2hat(i) + 7hat(j) + 4hat(k))m " to" vec(r )_(2)= (5 hat(i) + 2hat(j) + 8hat(k))m. The work done by the force is, |
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Answer» 18J |
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| 43. |
The time taken by a liquid to cool from 65^(@)C to 55^(@)C is 5 minutes and cools to 47^(@)C in the next 5 minutes, calculate the room temperature and the temperature of the body after another 5 minutes. |
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Answer» Solution :`log((theta_(2)-theta_(0))/(theta_(1)-theta_(0)))=(-K)/(mc)t=-K.T` where `K.=K/(mc)` is a constant `theta_(0)=65^(@)C,theta_(0)=55^(@)C,t=5 "minutes", theta_(0)=?` `"log"((55-theta_(0))/(65-theta_(0)))=K.xx5` …(i) For second part `theta_(1)=55^(@)C, theta_(2)=47^(@)Candt=5"minutes"` `"log"((47-theta_(0))/(55-theta_(0)))=-K.xx5` From the EQUATION (i) and (II) `"log"((55-theta_(0))/(65-theta_(0)))="log"((47-theta_(0))/(55-theta_(0)))` `(55-theta_(0))/(65-theta_(0))=(47-theta_(0))/(55-theta_(0))` `(55-theta_(0))^(2)=(65-theta_(0))(47-theta_(0))` `55^(2)-2xx55xxtheta_(0)+theta_(0)^(2)=65xx47-65theta_(0)-47theta_(0)+theta_(0)^(2)` `3025-110theta_(0)=3055-112theta_(0)` `theta_(0)=15^(@)C` LET `theta` be the temperature after another 5 minutes `"log"(theta-15)/(47-15)="log"((47-15)/(55-15))` `(theta-15)/32=32/(40)` `theta=40.6^(@)C` ANOTHER METHOD From Newton.s law of cooling Rate of cooling `prop`{MEAN difference of temperature between the body and the surroundings `(65-55)/(5) prop((54+55)/2)-theta_(0)` `2 prop 60-theta_(0)` ...(i) `(55-47)/5prop((55+47)/2)-theta_(0)` ...(ii) `8/(5)prop51-theta_(0)` dividing equation (i) and (ii) `(2xx5)/8=(60-theta_(0))/(51-theta_(0))` `510-10theta_(0)=480-8theta_(0)` `theta_(0)=30/(2)=15^(@)C` let `theta` be the temperature after another 5 minutes `(47-theta)/5prop((47+theta)/2)-15` ...(iii) from equation (i)`2prop60-theta_(0)` `2prop60-15` `2prop45` ...(iv) Dividing equation (iii) and (iv) `(47-theta )/(5xx2)=((47+theta)/2-15)/45` `47-theta/10=47/(2xx45)+theta/(2xx45)-15/(45)` Solving,`theta=40.6^(@)C` |
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| 44. |
The blades of an aeroplane propeller are 2 m long and rotate at 300 rpm. Calculate (i) the frequency, (ii) period of rotation, (iii) the angular velocity, (iv) linear velocity of point 0.5 m from the top of the blades. |
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Answer» SOLUTION :Length of the blade = r = 2M Frequency = `n = 300 RPM = 300/60 = 5 rps` Distance of the point = s=2-0.5 = 1.5 m Period `T=1/n = 1/5 = 0.2 s` ANGULAR velocity `=omega = 2pin = 2 xx 3.14 xx 5 = 31.4` rad/s Linear velocity = `v = somega = 1.5 xx 31.4 = 47.1` m/s |
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| 45. |
Two planets if radii in the ratio 2:3 are made from the material of density in the ratio 3:2. Then the ratio of acceleration due to gravity at the surface of the two planets will be ........... |
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Answer» `1:1` `g_1/g_2=(R_1rho_1)/(R_2rho_2)=((R_1)/(R_2))((rho_1)/(rho_2))=(2/3)(3/2)=1/1` `g_1/g_2 = 1:1` |
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| 46. |
A person drives a car along a circular track on a level ground. Why do we give banking to crurved roads ? |
| Answer» Solution :By doing so, the component of normal reaction along the center of the arc also contributes to the CENTRIPETAL FORCE and the MAXIMUM safe SPEED on the track is increased. | |
| 47. |
The equation of a stationary wave is y=10 sin((pix)/4)cos20pit.Distance between two consecutive nodes is |
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Answer» 4 |
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| 48. |
Two blocks A and B of masses m and 2m respectively are placed on a smooth floor. They are connected by a spring. A third block C of mass m moves with a velocity V, along the line joining A and B and collides elastically with A, as shown in figure. At a certain instant of time t, after collision, it is found that the instantaneous velocities of A and B are the same. Further, at this instant the compression of the spring is found to be x_(0). Determine (i) the common velocity of A and B at time t_(0)and (ii) the spring constant. |
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Answer» |
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| 49. |
(a) In a frame of reference S_(1) though the net force is zero, the net acceleration is not zero.(b) In a frame of reference S_(2), through the net force is not zero, the net acceleration is zero.(c ) In a frame of reference S_(3), the net acceleration is zero whenever the net force is zero. |
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Answer» `S_(1)` and `S_(3)` are INERTIAL and `S_(3)` is non - incrtial |
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| 50. |
A body is projected at angle 30^(@) to horizontal with a velocity 50 ms^(-1). Its time of flight is (g = 10 m//s^(2)) |
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Answer» 4s |
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