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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. |
If a bullet of mass 5 g moving with avelocityof 100m//s penetrates a wooden block upto 6cm . Find the average force imposed. |
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Answer» Solution :`u=100 m//s , v=0, S = 6XX10^(-2)m, a=? , v^(2)-u^(2)=2as` `0^(2)-(100)^(2)=2xxaxx6xx10^(-2)` `a=(-100xx100)/(2xx6xx10^(-2)),a=-1/(12)xx10^(6) m//s` `F=ma =5xx10^(-3)xx[-1/(12)xx10^(6)],F= (-5000)/(12)=-417 n` Retarding forceF=417 N |
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| 2. |
The cross-sectional area of the U-tube shown in the figure is everywhere uniform and of value 1.25 xx 10^(-3)m^(2). The horizontal section of the tube is of length 20 cm. When at rest, the limbs of the tube contain a liquid of density 2.5 up to equal heights. If the tube is rotated with angular velocity of 8.4 rad//s about one limb, calculate the volume of liquid that flows from one limb to the other. |
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Answer» |
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| 3. |
The angular velocity of the second's-hand ofa watch is |
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Answer» `(pi)/(15)rads^(-1)` |
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| 4. |
The unit of impulse is same as that of |
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Answer» energy |
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| 5. |
The thermodynamic process in which no work is done on or by the gas is |
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Answer» isothermal PROCESS `:.` WORK DONE, `W = P Delta V` = zero |
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| 6. |
Calculate the kinetic energy of a hollow metal sphere of mass 2 kg, which rolls over a horizontal surface with a linear velocity of 20 cm/s ? |
| Answer» SOLUTION :Toal K.E. =`(1//2)IOMEGA^(2) =(1//2)mv^(2) +(1//2) (2//3( MR^(2) XX (v^(2)//R^(2)) =(5//6) mv^(2) = 0.067 J` | |
| 7. |
Weightlessness experienced by spaceman in a space-ship orbiting the earth is the result of |
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Answer» inertia |
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| 8. |
(A): Thermodynamics process in nature are irreversible. (R ): Dissipative effects can not be eliminated. |
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Answer» Both (A) and(R ) are TRUE and (R ) is the CORRECT EXPLANATION of (A) |
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| 9. |
Write rule for error in result due to multiplication and division |
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Answer» Solution :Suppose two physical quantities A and B have measured values `A+-DELTAA,B +-DeltaB` respectively. Where `DeltaA and DeltaB` are their absolute errors. For a PRODUCT : Suppose the product of A and B is Z and the absolute error in Z is `DeltaZ`. `:.Z= AB` `:.Z+-DeltaZ=(A+-DeltaA)(B+-DeltaB)` `:.Z+DeltaZ=AB+-DeltaB+-BDeltaA+_(DeltaA DeltaB)` Dividing the left side by 2 and right side by AB, `1+-(DeltaZ)/(Z)=1+(DeltaA)/(A)+-(DeltaB)/(B)+-((DeltaA)/(A))((DeltaB)/(B))` `(DeltaA)/(A) and (DeltaB)/(B)` are very small hence their product will be very small and can be neglected. Neglecting negative sign for maximum error `(DeltaZ)/(Z)=(DeltaA)/(A)+(DeltaB)/(B)` For a Quotient : Suppose the quotient of A and B is Z and the absolute error of Z is `DeltaZ` `:.Z=(A)/(B)` `:.Z+DeltaZ=(A+-DeltaA)/(B+-DeltaB)` `Z(1+-(DeltaZ)/(Z))==(A(1+-(DeltaA)/(A)))/(B(1+-(DeltaB)/(B)))` `(1+(DeltaB))/(B))^(-1)` writing first two terms of binomial expansion, `(1+-(DeltaB)/(B))^(-1)=(1-+-(DeltaB)/(B))` From EQUATION (1), `:.1+-(DeltaZ)/(Z)=(1+(DeltaA)/(A)) (1+-(DeltaB)/(B))` `:.1+-(DeltaZ)/(Z)=1+(DeltaA)/(A)+-(DeltaB)/(B)+-((DeltaA)/(A))((DeltaB)/(B))` But `((DeltaA)/(A)) and ((DeltaB)/(B))` are smaller, so their product will be zero by ignoring them, `+(DeltaZ)/(Z)=+-(DeltaA)/(A)+-(DeltaB)/(B)` For relaive errros in Z, ignoring, regative signs `(DeltaZ)/(Z)=(DeltaA)/(A)+(DeltaB)/(B)` Hence the rule : When two quantities are multiplied or divided, the relative errorin the RESULT is the sm of the relative errors in the multipliers. |
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| 10. |
Assertion: If the observer the source, then the apparent frquency is f'= [(gamma + gamma_(0))/(gamma + gamma_(s))]f Reason: Doppler effect in sound is asymmetrical whereas Doppler effect in light is symmetrical. |
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Answer» Assertion and REASON are CORRECT and Reason is the correct EXPLANATION of Assertion |
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| 11. |
The principle used by a gymnast to increase the number of somersaults, is law of conservation of |
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Answer» LINEAR momentum |
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| 12. |
A person measures the time period of a simple pendulum inside a stationary lift and finds it to be "T". If the lift starts accelerating upwards with an acceleration y/3, the time period of the pendulum will be |
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Answer» `sqrt(3)T` |
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| 13. |
Even though electric flux is a scalar quantity, we consider the flux flowing out of a surface as positiveand flux entering into the surface as negative. Keeping this fact in mind answer the following question: The total electric flux through a Gaussian surface is zero when there is a charge outside the surface. Why is it so when the charge produces electic field? |
| Answer» Solution :The electric FIELD will always be along the DIRECTION away from the charge (assumed POSITIVE). But the flux will have both positive and negative values with equal magnitudes. HENCE total flux will be zero for charge outside the closed surface. | |
| 14. |
What is significant number? What is significant digit? |
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Answer» Solution :EVERY measurement involve errors. The result of measurement should be reported in a way that indicates the precision of measurement. Normally the reported result of the measurement is a number that includes all digit in the number that are known reliably plus the first digit that is uncertain. The reliable digits plus the first uncertain digits are known as significant digits or significant figures. The digits of significant figure are called significant digit. In any significant figure the last digit (the right most) is uncertain. For example, periodic time of simple pendulum is 1.62 SEC. In this 1 and 6 are reliable and certain while digit 2 is uncertain. Thus in this observation there are three significant digit. When length of object reported aller measurement of 287.5 cm in this measurement 2, 8, 7 are certain while digit 5 is uncertain. Reporting the result of a measurement that includes more digit than significant digits is superfluous and misleading since it would give a wrong idea about precision of measurement. A choice of change of different units does not change the number of significant digit or figure in measurement. For example, in measurement of length 2.308 cm there are 4 significant digit. It can be represented as 0.02308 m or 23.08 mm or `23080 mum`. In each measurement significant digit are 2,3,0, 8 hence significant digit are 4. Thus, location of DECIMAL point is of no CONSEQUENCE in determining the number of significant figure. |
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| 15. |
Statements : (a) Centre of mass is that fixed point of a system of particles or a rigid body with in the bondaries of the system where the entire mass is concentrated (b) Centre of gravity of a body is the point through which the whole weight of the body acts. |
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Answer» Both a and B are TRUE |
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| 16. |
What is the relation between the velocity and temperature? |
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Answer» Solution :Let us consider an ideal gas WHOSE equation of state is PV = nRT …………(1) where, P is PRESSURE, V is volume, T is temperature, n is number of mole and R is universal gas constant. For a given mass of a molecules, equation (1) can be WRITTEN as `(PV)/T`= Constant ...............(2) For a FIXED mass m, density of the gas inversely varies with volume, i.e., `rho propto 1/V, V=m/rho`........(3) Substituting equation (3) in equation (2), we get `P/rho = cT`..........(4) where c is constant. The speed of sound in air given in equation `v_(A) = sqrt((B_(A)/rho)) = sqrt((gamma P)/rho) = sqrt(gamma v_(T))` cna be written as `v= sqrt((gamma P)/rho) = sqrt(gamma cT)`............(5) SINCE `v propto sqrt(T)`, the speed of sound varies directly to the square root of temperature in kelvin. Let `v_(0)` be the speed of sound at temperature at `0^(@)` C or 273 K and v be the speed of sound at any arbitrary temperature T (in kelvin), then `v/v_(0) = sqrt(T/273) = sqrt((273 +t)/273)` `v=v_(0)sqrt(1+t/273) -=v_(0)(1+t/546)` (using binomial expansion). Since, `v_(0) = 331 ms^(-1)` at `0^(@)` C, v at any temperature in `t^(@)C` is `v=(3310.60t)ms^(-1)` Thus, the speed of sound in air increases by `0.61 ms^(-1)` per degree celcius rise in temperature. Note that when the temperature is increased, the molecules will vibrate faster due to gain in thermal energy and hence, speed of sound increases. |
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| 17. |
State whether the following statement is true or false. Give very brief reason in support of your answer. The quantity e^(2)/(2 epsi_(0) hc) is dimensionless. Here e, h and c are electronic charge, Planck's constant and velocity of light respectively and epsi_(0) is the permittivity constant of free space. |
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Answer» so `[e^(2)/(2epsi_(0))]=[Fr^(2)]`, i.e., `e^(2)/(2epsi_(0)ch)=[(Fr^(2))/(ch)] rarr [M^(0)L^(0)T^(0)]` |
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| 18. |
A uniform disc rotating freely about a vertical axis makes 90 revolutions per minute. A small piece of wax of mass m gram falls vertically on the disc and sticks to it at a distance of r cm from the axis. If the number of rotations per minute reduces to 60, find the moment of inertia of the disc. |
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Answer» |
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| 19. |
Two solid spheres of same material but diameters in the ratio of 5 : 4 are at temperatures 227^(@)Cand127^(@)C respectively. The temperature of the surrounding is 27^(@)C and Stefan's law holds. Calculate the ratio of rates of loss of heat of the two spheres? |
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Answer» Solution :If `E_("net")` is the energy lost per second by UNIT SURFACE area, then `E_("net")=sigma(T^(4)-T_(0)^(4))` `therefore` Energy lost by the whole sphere in 1 sec = `4pir^(2)sigma(T^(4)-T_(0)^(4))` Now `("Rate of heat lost by the I sphere")/("Rate of heat lost by the II sphere")` = `(r_(1)^(2)(T_(1)^(4)-T_(0)^(4)))/(r_(2)^(2)(T_(2)^(4)-T_(0)^(4)))=(5/4)^(2)[(500^(4)-300^(4))/(400^(4)-300^(4))]=34/7` |
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| 20. |
A horizontal tube of 1m is closed at both ends. A mercury column of 20 cm is contained in the middle of the tube and two equal lengths of this tube contain air at an atmospheric pressure of 76cm. The tube is now placed vertically. Calculate the distance through which the mercury column will be displaced at a constant temperature. |
Answer» Solution : Let M and N are two parts of tube at the LEFT and right sides of O. Case (i) : When tube is horizontal, Lengths of M and N =40cm Pressure of M and=76 cm of Hg Case (II): Let the tube is inverted and placed in vertical position. Let the mercury column is displaced by a distance y by placing tube vertically  Lengths of M and N are (40-y) and (40+y) respecively. `P_(1)`= Preessure of par M `P_(2)`= Pressure of PART N As TEMPERATURE is constant Therefore, we can write PV=constant For M: `P_(0) (40 A)=P_(1) (40-y) A` `rArr P_(1)=(P_(0) (40))/((40-y)) .......(i)` For N: `P_(0) (40 A)=P_(2) (40+y) A` `rArr P_(2)=(P_(0)40)/((40+y)) ...........(ii)` Here A= area of cross section of tube Pressure at lower face `(P_(1))="Pressure at upper face "(P_(2))+` Pressure due to mercury column of height 20 cm. `P_(1)=P_(2)=20pg ........(iii)` Use (i) and (ii) in (iii) `P_(0)(40)/((40-y)) =(P_(0)40)/((40+y)) +20pg` `(40P_(0))/((40-y)) -(40P_(0))/((40+y)) =20pg` `40 XX 76 pg [(2y)/(1600-y^(2))]=20pg` `(52) (2y)=1600-y^(2)` `304 y=1600-y^(2)` `y^(2)+304 y-1600=0` Solving for y, we get y=5.18 cm |
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| 21. |
A: A body is thrown with a velocity u inclined to the horizontal at an angle theta (theta gt 0, 90^(@)). At the highest point, the angle between instantaneous velocity and acceleration is zero R: At the highest point velocity of the projectile is zero. |
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Answer» If both A and B are TRUE and R is the correct EXPLANATION of A |
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| 22. |
Show that the torque acting on a body is equal to the rate of change of angular momentum of the body. |
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Answer» Solution :We know that ANGULAR momentum , `vecL=vecrxxvecpor,vecL=vecrxxmvecv` `:." "(dvecL)/(DT)=(dvecr)/(dt)xxmvecv+vecrxx(d)/(dt)(mvecv)` `=vecvxxmvecv+vecrxxvecF=vecrxxvecF" "[becausevecvxxvecv=0]` =`vectau` So, the rate if CHANGE of angular momentum of a BODY is equal to the torque acting on the body. |
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| 23. |
A particle executes simple harmonic motion of amplitude A. The distance from the mean position where its kinetic energy is equal to its potential energy, is |
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Answer» 0.81 A |
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| 24. |
The torque acting on a particle about an arbitrary origin is zero, what can you say about its angular momentum' ? |
Answer» Solution :About the C.M. `tau_(0)=rF`. About the ORIGIN `0= tau_(0)+ (vecR XX vecF) = rF-rF=0`  L about the origin`=I_(0)omega + vecR xx mvecv= I_(0)omega - rmv` |
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| 25. |
A double star is a system of two stars rotating about their centre of mass only under their mutual gravitational attraction. Let the stars have masses m and 2m and let their separation be 1. Their time period of rotation about their centre of mass will be proportional to a) 1^(3//2) b) a c) m d) m^(-1//2) |
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Answer» only a & C are TRUE |
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| 26. |
Rocket propulsion is associated with |
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Answer» The CONSERVATION of the angular momentum |
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| 27. |
Two moles of an ideal gas with (C_(P))/(C_(V))= (5)/(3) are mixed with 3 moles of another ideal gas with (C_(P))/(C_(V))= (4)/(3). The value of (C_(P))/(C_(V)) for the mixture is |
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Answer» <P>1.38 and for first gas `(C_(P_(1)))/(C_(V_(1)))= (5)/(3) rArr C_(P_(1)) = (5)/(3) C_(V_(1))` Now `C_(P_(1))- C_(V_(1))= R` `:. (5)/(3) C_(V_(1))- C_(V_(1))= R` `:. 5C_(V_(1))- 3C_(V_(1)) = 3R` `:. 2C_(V_(1))= 3R` `:. C_(V_(1)) = (3R)/(2)` and `C_(P_(1)) = C_(V_(1)) + R` `=(3R)/(2) + R` `:. C_(P_(1)) = (5R)/(2)` Now for second gas `(C_(P_(2)))/(C_(V_(2))) = (4)/(3)` `:. C_(P_(2)) = (4)/(3) C_(V_(2))` and `C_(P_(2))- C_(V_(2)) R` `:. (4)/(3) C_(V_(2))-C_(V_(2))R` `:. 4C_(V_(2)) -3C_(V_(2)) = 3R` `:. C_(V_(2)) = 3R` and `C_(P_(2)) = C_(V_(2)) + R` `=3R + R` `:. C_(P_(2)) = 4R` `:.` For mixture `(C_(P))/(C_(V)) = (n_(1)C_(P_(1)) + n_(2) C_(P_(2)))/(n_(1) C_(V_(1)) + n_(2) C_(V_(2)))` `n_(1)=2` mole, `n_(2)=3` mole `:. (C_(P))/(C_(V)) = (2 xx (5R)/(2) + 3 xx 4R)/(2 xx (3R)/(2) + 3 xx 3R)` `=(5R + 12R)/(3R + 9R) = (17R)/(12R)` `=(17)/(12)= 1.4166` `=1.42` |
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| 28. |
When 10 N force act on a body, 1ms^-2 acceleration is produced on it. Mass of body will be …... |
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Answer» F = ma `:. m=F/a=10/1` = 10 KG |
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| 29. |
A circular metal plate of uniform thickness has a radius of 10 cm. A hole of radius 4 cm is punched on the plate a little away from its centre. Centres of the plate and of the hole are 5 cm apart. Find the centre of gravity of the plate with the hole. |
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Answer» Solution :Centre of gravity of the plate before the hole was punched was at its centre G Fig. The centre of gravity of the CIRCULAR part before the PUNCHING was at its centre `G_(1)`. Suppose the centre of gravity of the plate with the hole is at `G_(2)` on the line `GG_(1)`. Let `W_(1)` = weight of the plate with hole, `W_(2)` = weight of the disc that is cut off. `:. " " W_(1)xxGG_(2)=W_(2)xxGG_(1)" ""or",GG_(2)=(W_(2))/(W_(1))xxGG_(1)` But `GG_(1)=5 CM, W_(1)= pi(10^(2)-4^(2))xxrho(rho` = weight per unit AREA) and `W_(2)=pixx4^(2)xxrho` `:." "GG_(2)=(4^(2))/(10^(2)-4^(2))XX5=(20)/(21)`cm Hence the centre of gravity of the system will be `(20)/(11)` cm to the left of the centre of the plate. 
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| 30. |
A particle is projected with velocity v at an angle of theta with horizontal. The average angular velocity of the particle from the point of projection to impact equals: |
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Answer» `(G COS theta)/(thetav)` |
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| 31. |
The coefficient of volume expansion of glycerine is 49 xx 10^(-5) K^(-1). Whatis the fractional change in its density for a 30^(@) C rise in temperature ? |
| Answer» SOLUTION :`0.0147 = 1.5 XX 10^(-2)` | |
| 32. |
(A): The hard boiled egg and raw egg can be distinguished on the basis of spinning of both. (R): The moment of inertia of hard boiled egg is more as compare to raw egg. |
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Answer» Both .A. and .R. are true and .R. is the CORRECT explanation of .A. |
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| 33. |
Block A of weight 100 N rests on a frictionless inclined plane of slope angle 30^(@) A flexible cord attached to A passes over a frictionless pulley and is connected to block B of weight W Find the weight W for which the system is in equilibrium . |
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Answer» Solution :As is clear from the system will be in equibrium, when force on block `A` , up the plane is equal to force on `A` down the inclined Plane `W = mg sin 0 = mg sin 30^(@) = 100 XX(1)/(2) = 50 N`  .
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| 34. |
A disc rotates freely with rotational kinetic energy E about a normal axis through centre. A ring having the same radius but double the mass of disc is now, gently placed on the disc. The new rotational kinetic energy of the system would be |
| Answer» ANSWER :C | |
| 35. |
If the earth stops rotating about its axis, then will be the change in the value of g at a place in the equatorial plane? |
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Answer» `45^@` |
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| 36. |
A sinusoidal wave istraveling on a string with speed 40 cm/s. The displacementof the particle of the string at x = 10 cm varies with time according to y = (5.0cm) sin(1.0-4.0s^(-1))t)The linear density of the string is 4.0 g/cm |
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Answer» the frequency of wave is 0.64 Hz |
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| 37. |
If mu_k is the coefficient of kinetic friction, mu_l is the coefficient of rolling friction and us is the coefficient of static friction, then |
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Answer» `mu_sgt mu_k GT mu_r` |
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| 38. |
What is critical velocity ? Write its equation. |
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Answer» SOLUTION :The FLOW of FLUID in pipe remains streamline by a maximum velocity then this velocity is known as critical velocity. `v_(C)=(R_(E)eta)/(rhod)` |
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| 39. |
Suppose, the acceleration due to gravity at the Earth.s surface is 10 m//s^2and 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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| 40. |
A particle is subjected to a force which varies with distance as shown in figure . The work done on the particle at in dispalcement of 12 m is …… . . |
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Answer» 18 J  Area under the graph `F to d` indicates work done ` :. ` work done W = Area of ABCD + Area of `DeltaDCE` ` = 2 xx 4 +1/2 xx 5 xx2 ` ` = 8+5` ` :. W= 13 J ` |
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| 41. |
The capillary tubes of radii of 0.2 cm and 0.4 cm are dipped in the same liquid. The ratio of the heights through which liquid will rise in tubes is |
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Answer» `1:2` |
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| 42. |
Two rods of same length made of brass and steel of equal cross-sectional area are joined end to end as shown in figure (a) and supported between two rigid vertical walls. Initially the rods are unstrained. If the temperature of system is raised by DeltaT, find the displacement of the junction of the rods. Given that the coefficients of linear expansion and Young.s modulus of brass and steel are alpha_(b), alpha_(s) (alpha_(b) gt alpha_(s)), y_(b) and y_(s) respectively. |
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Answer» SOLUTION :Here the initial lengths of the two rods are `l_(0)`and as `alpha_(b) gt alpha_(s)`,we can directly state that the junction of the two rods is displaced TOWARD right. Due to this brass rod is some what expanded but less as compared to free expansion and steel is OVERALL compressed due to the stress developed between the two rods. Figure (b) shows the final situation of rods at higher temperature `Deltat`. In figure, A is the initial position of junction and B is its final position. If only brass rod is present, expansion in it due to increase in temperature by `DeltaT` is `Deltal_(b) = alpha_(b)l_(0) DeltaT`which is equal to AC in figure b. But finally the rod is expanded by x, thus the ELASTIC compression in the rod due to stress between the two rods is BC, which is given as `BC = alpha_(b) l_(0)DeltaT-x ` Thus elastic strain in brass rod is `("Strain")_("brass") =(alpha_(b)l_(0)DeltaT-x)/(l_(0))` [As `alpha_(0)l_(0)DeltaT` is small] Similarly if steel rod along is there, it would have been expanded by `Deltal_(s)=alpha_(s)l_(0)DeltaT` which is equal to AD in figure b. But finally steel rod is compressed by x, thus the elastic compression in steelrod due to stress between the two rods in BD, which is given as `BD=alpha_(s)l_(0)DeltaT +x` Thus elastic strain in steel rod is `("Strain")_("steel") =(alpha_(s)l_(0)DeltaT +x)/(l_(0)) "" ` [As `alpha_(s)l_(0)DeltaT` is small] As the two rods are in contact, stress developed in the two rods must be equal so, we have `("Stress")_("brass") =("Stress")_("steel")` (or) `y_(b) xx ("Strain")_("brass") =y_(s) xx ("Strain")_("steel") or y_(b)[(alpha_(0)l_(0)DeltaT-x)/(l_(0))]=y_(s)[(alpha_(s) l_(0) DeltaT+x)/(l_(0))]` (or) `x=((y_(b)alpha_(b)-y_(s) alpha_(s)) l_(0) DeltaT)/((y_(b)+y_(s)))` |
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| 43. |
An ideal gas is taken through a cyclic thermodynamic process though four steps. The amounts of heat involved in these steps are Q_1 = 5960 J, Q_2 = 5585 J, Q_3 = - 2980 J and Q_A = 3635 J respectively. The corresponding works involved are w_1= 2200 J, W_2= -825 J, W_3= -1100 J, and W_4respectively (i) Find the value of W_4(ii) What is the efficiency of the cycle? |
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Answer» Solution :In a cyclic PROCESS `Delta `U = 0 , Net HEAT absorbed by the system, `Q=Q_1 +Q_2+Q_3 +Q_4= 5960 - 5585 - 2980 + 3645 = 1040J` Net work DONE, `W_1= W_2 +W_3 +W_4= 2200 -825-1100 + W_4= 275+ W_4` From 1 st law of thermodynamics, `Q = Delta U + W (or) 1040 = 0 + 275+W_4` ` therefore W_4 = 1040 - 275 = 765 J` Efficiency of cycle `=("workdone(W)")/(" Heat absorved"(Q_1+Q_4)) =( 275+ 465)/( 5960 +3645 )=(1040 )/( 9605 )` % efficiency `=( 1040 )/( 9605 )xx 100 = 10.83%` |
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| 44. |
A stone is projected vertically up to reach a maximum height .h.. The ratio of its kinetic to potential energies at a height (4h)/(5) will be |
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Answer» `5:4` |
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| 45. |
A toy is made of a rectangular wooden frame with four small wheels at its vertices. A tight fitting belt of negligible mass runs around the frame passing over the wheels. Mass of the completetoy is M. Now a small block of mass m is stuck at the top of the right vertical segment of the belt and the system is released. Height of the toy is h. Find the speed of the block when it is about to hit the ground. Assume no slipping anywhere and neglect the dimension of the wheels |
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Answer» |
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| 46. |
A solid sphere has a radius of 1.5 cm and a mass of 0.038 kg. Calculate the specific gravity or relative density of the sphere. |
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Answer» Solution :Radius of the sphere R = 1.5 cm, MASS m = 0.038 kg Volume of sphere `V = 4/3 pi R^2 = 4/3 XX (3.14) xx (1.5 xx 10^(-2))^(3) = 1.413 xx 10^(-5) m^(3)` Therefore, DENSITY `rho = m/V = (0.038 kg)/(1.413 xx 10^(-5) m^3) = 2690 kg m^(-3)` Hence, the SPECIFIC of the sphere `= (2690)/(1000) = 2.69` |
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| 47. |
Find the speed of two objects if, when they move uniformly towards each other they gte 4 m closer in every see and when they move uniformly in the same direction with their original speeds, they get 4m closer every 10 s |
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Answer» 1.8 m/s, 1.6 m/s |
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| 48. |
Find the position of instantaneous axis of rotation of rod of length L when its end, A moves with a velocity vhati and the rod rotates with a angular velocitybaromega=-(V)/(2L)hatk |
| Answer» Solution :At a distance 2L from the point A and HENCE LIES outside the ROD | |
| 49. |
Why are army troops not allowed to march in steps while crossing a bridge? |
| Answer» Solution :Because,the frequency of BOOT STEPS MAY match with NATURAL frequency of the bridge and due to resonance,the bridge may pick up LARGE amplitude and break. | |