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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 force of constant magnitude F_(0) is applied in the tangential direction as shown in the figure. Assume that the bob is at its lowest point initially. |
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Answer» SPEED of bob at `theta=60^(@) " is "sqrt((2L)/(m)[(F0pi)/(3)+(mg)/(2)])` |
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| 2. |
According to Stefan's law of radiation, a black body radiates energy sigmaT^(4) from its unit surface area every second where T is the surface temperature of the black body and sigma=5.67xx10^(-8)W//m^(2)K^(4) is known as Stefan's constant. A nuclear weapon may be thought of as a ball of radius 0.5 m. When detonated, it reaches temperature of 10^(6) K and can be treated as a black body. If all this energy U is in the form of radiation, corresponding momentum is p=U/c. How much momentum per unit time does it impart on unit area at a distance of 1 km ? |
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Answer» Solution :Momentum in unit TIME, `p=(U)/(c)` `=(1.8xx10^(17))/(3xx10^(8))` `=6xx10^(8)" N/s"` Momentum in unit time per unit area is PRESSURE, `:.P=(p)/(A)` `=(6xx10^(8))/(4pid^(2))` `=(6xx10^(8))/(4xx3.14xx(10^(3))^(2))` `=0.4777xx10^(2)` `:.P=47.77" Nm"^(-2)` |
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| 3. |
A) When a body slides on rough surface, its momentum is not conserved. B) When a ball falls from a height, momentum of earth ball system is conserved C) In case of explosion of flying bomb, momentum is conserved. |
| Answer» Answer :A | |
| 4. |
According to Stefan's law of radiation, a black body radiates energy sigmaT^(4) from its unit surface area every second where T is the surface temperature of the black body and sigma=5.67xx10^(-8)W//m^(2)K^(4) is known as Stefan's constant. A nuclear weapon may be thought of as a ball of radius 0.5 m. When detonated, it reaches temperature of 10^(6) K and can be treated as a black body. If surrounding has water at 30^(@)C, how much water can 10% of the energy produced evaporate in 1 s ? [S_(w)=4186.0" J/kg K and "L_(v)=22.6xx10^(5)" J/kg"]. |
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Answer» Solution :Energy obtained at each SECOND, `U=1.8xx10^(17)" J/s"` Heat REQUIRED to convert WATER into steam at each second, `Q=10%` of U `=1.8xx10^(17)xx(10)/(100)` `=1.8xx10^(16)" J/s"` Heat used in heating water of .m. mass from `30^(@)C` to `100^(@)C` and then to convert it into steam in each second, `Q.="mS"_(w)Deltatheta+mL_(v)` `=mxx4186xx(100-30)+mxx22.6xx10^(5)` `Q.=2.93xx10^(5)m+22.6xx10^(5)m` `Q.=25.53xx10^(5)m` According to question, `Q=Q.` `1.8xx10^(16)=25.53xx10^(5)m` `:.m=(1.8xx10^(16))/(25.53xx10^(5))` `:.m=0.07050xx10^(11)` `:.m~~7.0xx10^(9)` kg |
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| 5. |
According to Stefan's law of radiation, a black body radiates energy sigmaT^(4) from its unit surface area every second where T is the surface temperature of the black body and sigma=5.67xx10^(-8)W//m^(2)K^(4) is known as Stefan's constant. A nuclear weapon may be thought of as a ball of radius 0.5 m. When detonated, it reaches temperature of 10^(6) K and can be treated as a black body. Estimate the power it radiates. |
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Answer» Solution :Power emitted according to STEFAN - Boltzmaan LAW, `P=sigmaAT^(4)` `=SIGMA(4piR^(2))T^(4)` `=5.67xx10^(-8)xx4xx3.14xx(0.5)^(2)xx(10^(6))^(4)` `=17.8xx10^(18)` `~~1.8xx10^(17)" J/s"` |
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| 6. |
A box of mass 8 Kg is placed on a rough inclined plane of inclination q. Its downward motion can be prevented by applying an upward pull F. And it can be made to slide upwards by applying a force 2F. The coefficient of friction between the box and inclined plane is |
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Answer» `1/3 TAN THETA ` |
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| 7. |
A 2kg block is placed over a 4kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0.20. The acceleration of the two blocks if a horizontal force of 12N is applied to the lower block is |
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Answer» `2ms^(-2) , 2ms^(-2)` |
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| 8. |
Steel rod at 40^(@)C is observed to be 1 m long when measured by a brass scale which is correct at 0^(@)C. correct length of the steel rod at 0^(@)C is (alpha_("steel")=12xx10^(-6)//1^(@)Candalpha_("Brass")=19xx10^(-6)//1^(@)C) |
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Answer» 99.89 cm |
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| 9. |
(A) The focal length of lens changes when red light is replaced by blue light ( R) The focal length of lens does not depend on colour of light speed. |
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Answer» Both A and R are TRUE and R is the correct explanation of A |
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| 10. |
Assertion : A body is first taken along a smooth inclined plane through certain height . In the next case it is lifted vertically upwards to same height , workdone by gravity in both cases is same Reason : Gravitational force is conservative force |
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Answer» Both ASSERTION and REASON are true and Reason is the correct EXPLANATION of Assertion |
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| 11. |
"The cap of a pen can easily be opened with the help of two fingers than with one finger". Explain Why? |
| Answer» Solution :The two FORCES APPLIED by the fingers FORM a couple. The moment of this couple will be greater than the TORQUE applied by the SINGLE force. | |
| 12. |
The pressure and the temperature of an ideal gas in an adiabatic process are related as p prop T^(3). What is the value of the ratio C_(p)//C_(v) of the gas? |
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Answer» Solution :For 1 mol of an ideal gas, `pV = RT or, T = (pV)/(R )` Now, `p prop T^(3)` or, `p = kT^(3) = k((pV)/(R ))^(2) [k = "constant"]` `:. p = k(p^(3)V^(3))/(R^3) or, p^(2)V^(3) = (R^3)/k` or, `pV^(3//2) = ((R^3)/(k))^(1//2)` or, `pV^(3//2) = "constant"` For an ADIABATIC process, `pV^(gamma)` = constant, So, `gamm = (C_p)/(C_v) = 3/2` |
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| 13. |
A plane mirror is placed along the y-axis such that x-axis is normal to the plane of the mirror. The reflecting surface of the mirror is towards negative x-axis. The mirror moves in positive x-direction with uniform speed of 5 m/s and a point object P is moving with constant speed 3m/s in negative x-direction. find the speed of image with respect to mirror in m/s. |
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Answer» |
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| 14. |
Obtain an equation of continuity for a flow of fluid on the basis of conservation of mass. |
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Answer» Solution :The MASS flow RATE through a pipe, it is is necessary to assume that the flow of fluid is steady, the flow of the fluid is said to be steady if at any given point, the velocity of each passing fluid particle remains constnat with respect to time. Under this condition, the path taken by the fluid particle is a streamline. Consider a pipe AB of varying cross sectional area `a_1 and a_2` such that `a_1 > a_2`. A non-viscous and incompressible liquid FLOWS steadily through the pipe, with velocities `v_1 and v_2` in area `a_1` and`a_2`, respectively. Let `m_1` be the mass of fluid flowing through section A in time `Deltat, m_1 = (a_1v_1Deltat) rho` Let `m_2` be the mass of fluid flowing through section B in time `Deltat, m_2 = (a_2v_2 Deltat) rho` For an imcompressible liquid, mass is conserved `m_1 = m_2` `a_1v_1 Deltat rho = a_2v_2 Deltat rho` `a_1 v_1 = a_2v_2 implies av = "constant"` Which is called the equation of continuity and it is a statement of conservation of mass in the flow of FLUIDS. In general, a v = constant, which means that the volume flux rate remains constant throughout the pipe. In other words , the smaller the cross section, greater will be the velocity of the fluid. 
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| 15. |
Statement (a) : during simple harmonic oscillation kinetic energy converted in potential energy and vice-versa Statement (b) : Total mechanical energy of simple harmonic oscillator is directly proportional to sqaure of the frequency of oscillation Statement (c) : Simple harmonic oscillator obeys the law of conservation of energy. Statement (d) : Total mechanical energy of oscillator is directly proportional to a square of the amplitude of the oscillation |
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Answer» a, B are true |
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| 16. |
A body of mass 10 kg is placed on rough surface, pulled by force F making an angle of 30^(@) above the horizontal. If the angle of friction is also 30^(@), then the minimum magnitude of force F required to move the body is equal to( take, g=10 ms^(-2)) |
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Answer» 100 N  `N=mg-F-sin 30^(@)=(100-(F)/(2))` ` implies"" (sqrt(3)F)/(2)=(1)/(sqrt(3))(100-(F)/(2))` ` implies (3F)/(2) =100-(F)/(2) implies F=50N` |
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| 17. |
In a sonometer, the waves produced are |
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Answer» PROGRESSIVE LONGITUDINAL waves |
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| 18. |
Four thin uniform rods each of mass m and length l are arranged to form a square. Find the moment of inertia of the system about an axis (i) Passing through its centre and perpendicular to its plne. (ii) Passing through one of its sides. (iii) Passing through a corner and perpendicular to its plane. (iv) About a diagonal of the system. |
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Answer» Solution :(i) M.I. about an axis PASSING through its centre and perpendicular to its PLANE : `4[(ml^(2))/(12)+m((l)/(2))^(2)]=(4ml^(2))/(12)xx4` (II) `I_(s)=(ml^(2))/(3)xx2+ml^(2)` `=(5)/(3)ml^(2)` (iii) From parallel axes theorem `I=I_(C )+mr^(2)=(4ml^(2))/(3)+4m((l)/(sqrt(2)))^(2)=(10)/(3)ml^(2)` (iv) From perpendicular axes theorem `I_(z)=I_(X)+I_(y) "" I_(z)=2I_(x)` `I_(x)=(I_(z))/(2)=(I_(C ))/(2)=(2)/(3)ml^(2)` 
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| 19. |
(A) : A unitless physical quantity must be dimensionless. (R) : A pure number is always dimensionless. |
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Answer» Both (A) and (R) are true and (R) is the CORRECT explanation of (A) |
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| 20. |
A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diameter of the rim and the iron ring are 5.243m and 5.231 m respectively at 27" "^(@)C. To what temperature should the ring be heated so as to fit the rim of the wheel ? (alpha_(1)=1.20xx10^(-5)K^(-1)) |
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Answer» Solution :Given, `T_(1) = 27^(@) C` ` L_(T 1) = 5.231 m` ` L_(T2) = 5.243 m` So, ` L_(T2) = L_(T1) [1 + alpha_(1) (T_(2) - T_(1)]` ` 5.243 m = 5.231 m [1 + 1.20 xx 10^(-5) K^(-1) (T_(2) - 27^(@) C)]` or `T_(2) = 218^(@) C`. |
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| 21. |
The pendulum of a certain clock has time period 2.04s. How fast or slow does the clock run during 12 hour ? |
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Answer» 11.4 MINUTES slow |
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| 22. |
The Young's modulus of a material having a cross-sectional area of1cm^(2)is 2 xx 10^(12)"dyne/cm"^(2).If the length is to be doubled the force required is |
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Answer» `4XX10^(7)N` |
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| 23. |
An important method in order to measure large distances such as the distance of a planet or a star from the earth is |
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Answer» the PARALLEL method |
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| 24. |
A body is projected with an angle theta. The maximum height reached is h. If the time of flight is 4 sec and g = 10m//s^2 , the value of, then the value of h is |
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Answer» 40m |
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| 25. |
Whatis acyclicprocess? |
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Answer» Solution :This is a thermodynamic process in which the thermodynamic system returns to its initial state after undergoing a series of changes. SINCE the system comes back to the initial state, the change in the internal energy is zero. In CYCLIC process, heat can flow in to system and heat flow out of the system. From the first law of THERMODYNAMICS, the net heat transferred to the system is equal to work DONE by the gas. `Q_("net") = Q_("in") - Q_("out") = W" (for a cyclic process)"` |
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| 26. |
A spring is connected to a mass m suspended from it and its time period for vertical oscillation is T. The spring is now cut into two equal halves and the same mass is suspended from one of the halves . The period of vertical oscillation is |
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Answer» `T'=SQRT(2)T` |
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| 27. |
What will be the nature of change in internal energy in case of precesses shown below? |
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Answer» only a,B are CORRECT |
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| 28. |
A positive charge q is placed in front of a conducting solidcube at a distanced from its centre. Find theelectricfield at the centre of the cubedue tothe charge appearing on its surface. |
Answer» Solution :Charges will be inducedon the SURFACEOF thecube due to the charge q. Thenet electricfield at the centre of the cubedue to all the chargemustbe zero . Let `E_(1)` bethe electricfield due to chargeappearingon thesurface of the CUBE . If `E_(2)`is theelectric field due to charge q, then.  `vec(E_(1)) + vec(E_(2)) = 0 or vec(E_(1)) - vec(E_(2)) or vec(E_(1)) = vec(E_(2))` The electricfield at the centreof thecube, due tocharge q. `E_(2) = ((1)/(4 piin_(0))) (q)/(d^(2))` `rArr E_(1) = ((1)/(4pi in_(0))) (q)/(d^(2))` |
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| 29. |
A train is moving on a straight track with acceleration a. A passenger drops a stone. What is the acceleration of stone with respect to passenger? |
| Answer» Solution :`SQRT(a^(2)+g^(2))` where g = ACCELERATION DUE to GRAVITY. | |
| 30. |
A rod AB of uniform cross-section consists of 4 sections AC, CD, DE and EB of different metal with thermal conductivities K, 0.8,1.2K and 1.5 respectively. Their lengths are respectively L,1.2L, 1.5L and 0.6L. They are jointed rigidly in Succession at C, D and E to from the rod AB The end A is mainained at 100^(@)C and the end B is the maintained at 0^(@)C. The steady state temperatures p the joints C, D and E are respectively T_(C),T_(D) and T_(E) Match the following two columns |
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Answer» |
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| 31. |
{:(,"List - I",,"List - II"),((a),"Angular acceleration",(e ),J.S),((b),"Moment of inertia",(f),Kgm^(2)),((c ),"Torque",(g),rad//sec^(2)),((d),"Angular momentum",(h),N.m):} |
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Answer» a-g, b-f, c-h, d-e |
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| 32. |
What is the temperature of the steel-copper junction is the steady state of the system shown in figure. Length of the steel rod = 15.0 cm, length of the copper rod = 10.0 cm, temperature of the furnace =300^(@)C, temperature of the other end =0^(@)C. The area of cross section of the steel rod is twice that of the copper rod. (Thermal conductivity of steel =50.2" Js"^(-1)m^(-1)K^(-1), and copper =385" Js"^(-1)m^(-1)KJ^(-1)). |
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Answer» Solution :Temperature of furnance `T_(1)=300^(@)C` Temperature of ice box `T_(2)=0^(@)C` Temperature of junction `=T^(@)C` By taking sign 1 for steel and 2 for copper, Thermal CONDUCTIVITY of steel `K_(1)=50.2" Js"^(-1)m^(-1)K^(-1)`, length `L_(1)=15` cm Thermal conductivity of copper `K_(2)=385" Js"^(-1)K^(-1)`, length `L_(2)=10` cm HEAT current joining in steel and copper is same under thermal study state `H_(1)=H_(2)` `:.(K_(1)A_(1)(T_(1)-T))/(L_(1))=(K_(2)A_(2)(T-T_(2)))/(L_(2))` `:.(50.2xx2A_(2)(300-T))/(0.15)=(385xxA_(2)(T-0))/(0.10)` `:.(50.2xx2xx(300-T))/(3)=(385T)/(2)` `:.200.8(300-T)=1155T` `:.60240-200.8T=1155T` `:.60240=1355.8T` `:.T=(60240)/(1355.8)` `:.T=44.43^(@)C` |
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| 33. |
If two spheres of same material and same radius r are touching each other, then show that the gravitational force between them is directly proportional to r^4 |
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Answer» SOLUTION :`F = (Gm_(1) m_(2))/r^2` `=G/r^2 [(4/3pir^2rho)(4/3pir^2 rho)]` `=(4/9pi^2rho^2G)r^4` `:. F alphar^4` |
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| 34. |
The particle executing simple harmonic motion has a kinetic energy K_(0)cos^(2) omega t. The maximum value of potential energy and the total energy are respectively…………and………….. |
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Answer» `(K_0)/(2)" and "K_(0)` When KINETIC energy X is zero then its potential energy is maximum. Means `U_("max")+0=E`. Hence `U_("max")= E"""……."(1)` Kinetic energy `=(1)/(2) m OMEGA^(2) a^(2) cos^(2) omega t` `= K_(0) cos^(2) omega t` (is given) `therefore K_(0)= (1)/(2) m omega^(2) a^(2)=` Total mechanical energy Now, total mechanical energy `= K_(0)` and maximum potential energy `U_("max")= E= K_(0)`. |
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| 36. |
The error in measurement of radius of a sphere is 1%. What is the error in the measurement of volume? |
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Answer» Solution :`(DELTA r)/(r ) xx 100=1%` VOLUME of SPHERE `V= (4)/(3) pi r^(3)` Error in the MEASUREMENT of volume `(Delta V)/(V) xx 100= 3 (Delta r)/(r ) xx 100 = 3 xx 1= 3%` |
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| 37. |
A ship of mass 3xx10^(6) kg initially at rest is pulled by a force 6xx10^(4)N through a distance of 4m. The speed of the ship is (Assume resistivity of water is negligible) |
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Answer» 1.5 m/s Now, `v^2 - u^2 + 2AS` `v^2 = 0 + 2 XX 2 xx 10^(-2) xx 4` `v^2 = 16 xx 10^(-2) =0.16` `v=0.4 ms^(-1)` |
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| 38. |
A faulty thermometer has its fixed points marked as 5^(@) and 95^(@). The temperature of a body as measoured by the faulty thermometer is 59^(@). Find the correct temperture of the body on Celsius scale. |
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Answer» Solution :LET the reading on the FAULTY thermometer be x and the corresponding reading on Celsius SCALE be C. `(x -"L.F.P.")/("U.F.P. - L.F.P.") = (C - 0)/(100 - 0) "" (x - 5)/(95 - 5) = (C)/(100)` ` C = (100)/(90) (x -5 ) "" C = (10)/(9) (x - 5)` ` C = (10)/(9) (59-5) "" (because x = 59^(@)) C = 60^(@) C ` |
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| 39. |
In a harbour, wind is blowing at the speed of 72 km*h^(-1) and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction . If the boat starts moving at a speed of 51 km*h^(-1) to the north,what will be the direction of the flagon the mast of the boat ? |
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Answer» |
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| 40. |
Ice melts at the rate of 5 kg per hour in a cold storage when the temperature outside is 300 K. Calculate the minimum power of the motor is to be used to drive a refrigerator to just prevent the melting of ice. Specific latent heat of usion of ice =3.3 xx 10 ^(5) J kg ^(-1) .[E.Q] |
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Answer» SOLUTION :`T_(1) = 300 , T_(2) = 273 K` Quantity of heat to be removed from the SINK `= Q _(2) = (ML)/(t)` `Q_(2) = (% xx 3>3 xx 10 ^(5))/(60 xx 60) = 458.33 J//S` |
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| 41. |
The minimum velocity of projection of a body to send it to infinity from the surface of a planet is (1)/(sqrt(6)) times that is required from the surface of the earth. The radius of the planet is (1)/(36) times the radius of the earth. The planet is surrounded by an atmosphere which contains monoatomic innert gas (gamma = 5//3) of constant density up to a height h(h lt lt radius of the planet). Find the velocity of sound on the surface of the planet. |
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Answer» <P> Solution :Escape velocity from the surface of the planet`v_(p)=sqrt(2g_(p)R_(p))` Given `v_(p)=(v_(e ))/(sqrt(6)) = sqrt((2g_(e )R_(e ))/(6))` `sqrt((g_(e )R_(e ))/(3)) = sqrt(2g_(p)R_(e )//36) rArr g_(p) =6g_(e )` Pressure exerted by the atmospheric column of height h on the surface of the planet `P = rho g_(P) h` USING equation of state `P = (rho RT)/(M)` Hence SPEED of the sound `v = sqrt((yRT)/(M)) = sqrt(yg_(p)h) = sqrt(6y g_(e )h) = sqrt(10 g_(e )h)` |
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| 42. |
Derive Mayer's relation for an ideal gas. |
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Answer» Solution :(i) Molar specific HEAT capacity is defined as heat energy required to increase the temperature of one MOLE of substance by 1K or `1^(@)C` (ii) Mayer.s relation: Consider `mu` mole of an ideal gas in a container with volume V, pressure P and temperature T. When the gas is heated at constant volume the temperature increases by dT. As no work is done by the gas, the heat that flows into the system will increase only the internal energy. LET the change in internal energy be dU. If `C_(v)` is the molar specific heat capacity at constant volume, from equation. `C_(v) = (1)/(mu) (dU)/(dT)`...(1) dU = `muC_(v)dT`...(2) Suppose the gas is heated at constant pressure so that the temperature increases by dT. If .Q. is the heat supplied in this process and .dV. the change in volume of the gas. Q = `muC_(P)dT`...(3) If W is the workdone by the gas in this process, then W = PdV...(4) But from the first law of thermodynamics, Q = dU + W...(5) Substituting equations (2), (3) and (4) in (5), we get, `muC_(P)dT = muC_(v)dT + PdV` For mole of ideal gas, the equation of state is given by PV = `muRT rArr PdV + VďP = muRdT` Since the pressure is constant, dP = 0 `C_(P)dT = C_(V)dT + RdT` `:. C_(P) = C_(V) + R` (or) `C_(P) - C_(V) = R`...(6) This relation is called Mayer.s relation It implies that the molar specific heat capacity of an ideal gas at constant pressure is greater than molar specific heat capacity at constant volume. The relation shows that specific heat at constant pressure `(s_(P))` is always greater than specific heat at constant volume `(s_(v))`. |
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| 43. |
The lengths of smooth & rough inclined planes of 45^(@) inclination is same. Times of sliding of a body on two surfaces is t_(1), t_(2) and mu = 0.75, then t_(1):t_(2)= |
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Answer» `2:1` |
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| 44. |
(A): Metals have larger values of Young's modulus than its alloys (R): A material with larger value of Young's modulus requires a large force to produce small changes in its length |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A) |
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| 45. |
Mention any two physical significance of moment of inertia. |
| Answer» Solution :It PLAYS the same role in ROTATORY MOTION as the mass does in TRANSLATORY motion. | |
| 46. |
If vecAxxvecB =vecCxxvecD "and"vecAxxvecC=vecBxxvecD+"while"|vecA|-!=|vecD|,|vecB|!=|vecC|, show (vecA-vecD)that, and (vecB-vecC)are parallel. |
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Answer» Solution :Given, `vecAxxvecB=vecCxxvecD` `vecAxxvecC =vecBxxvecD` Subtracting (2)from (1), we get , `vecAxxvecB-vecAxxvecC=vecCxxvecD -vecBxxvecD` `:. vecAxx(vecB-vecC)` `=(vecC -vecB)xxvecD[:.vecPxx(vecQ+vecR)=vecPxxvecQ+vecPxxvecR]` `:. vecAxx(vecB-vecC)=-(vecB-vecC)xxvecD``:. vecAxx(vecB-vecC)+(vecB-vecC)xxvecD=0 ``:.vecAxx(vecB-vecC)-vecDxx(vecB-vecC)=0` `[:.PxxvecQ=-vecQxxvecP]` `:.(VECA-vecD)xx(vecB-vecC)=0` Now,`|vecA|!=|vecD|"and"|vecB|!=|vecC|` `:. (vecA-vecD)!=0"and" (vecB-vecC)!=0` `(vecA-vecD)"and"(vecB-vecC)` are at an angle `0^(@)`or `(vecA-vecD)"and"(vecB-vecC)` are parallel. `[:. vecPxxvecQ=PQsin THETA "and"P!=0!=Q."hence " sin theta=0]` |
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| 47. |
Four particle each of mass 1kg are at the four orners of a square of side lm. The workdone to move one of the particles to infinity: |
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Answer» `2sqrt(2)G` |
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| 48. |
In a criket match a batsman hits a six . Ball is hit 1 m above the ground level from a point which is 110m away from the benches of stadium . Ball leaves the bat with a speed 35 m//s and at an angle 53^(@) abovethe horizontal. Height and width of benches are 1 m each . Assume thatbenches are perpendicular to the plane of motion of the ball. Which bench will the ball hit ? |
Answer» Solution :Situation DESCRIBED in question is shown inthe following figure . Initial point of motion of ball is MARKED as point O and is selectedas the origin of the coordinate system . X-axis is horizontal and Y-axis is vertically upward.  Benches are 1 m high and 1 m wide and if we join the edge of the benches, then we get a planeinclined at an angle `45^(@)` with the horizontal . On thisimaginary palne , length of each step is `sqrt2m` . Let the ball hit on thisplane at a distnce x from its bottom . Components of displacement of ball can be written as follows : `S_(x)=110+x cos 45^(@)=110+(x)/(sqrt2),` `S_(y)x sin 45^(@)=(x)/(sqrt2),` Components of initial velocity are shown in the figure . `u_(x)=21m//s,u_(y)=28m//s,` We can write the components of acceleration as follows: `a_(x)=0,a_(y)=-9.8m//s^(2)`, X-axis, `S_(x)=u_(x)t+(1)/(2)u_(x)t^(2) rArr 110+(x)/(sqrt2)=21t` Y-axis : `S_(y)=u_(y)t+(1)/(2)a_(y)t^(2) rArr(x)/(sqrt2)=28t-4.9t^(2)` Substituting VALUE of `(x)/(sqrt2)` from equation (ii) in equation (i), we get the following: `rArr 110+28t-4.9t^(2)=21t` `rArr 4.9t^(2)-7t-110=0` `rArr t=(7pmsqrt(49+4xx4.9xx110))/(9.8)` `rArr t=(7 pm 7sqrt45)/(9.8)` We shall consider the positive sigh only and rearrange as follows: `rArr t=(1+3sqrt5)/(1.4)=5.5s` Substituting above result in equation (1), `rArr 110 +(x)/(sqrt2)=21xx5.5 rArr x=5.5 sqrt2` We have already discussed that each step is `sqrt2` m, HENCE `(x)/(sqrt2)=5.5` We can that it is more than `5^(t"th")` step , hence ball hits the `6^("th")` step. |
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| 49. |
The mass of a beaker is (10.1 pm 0.1)g when empty and (17.3pm0.1)g when filled with a liquid. The mass of the liquid with possible limits of accuracy is |
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Answer» `(7.2 PM 0.2)G` |
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| 50. |
Moment of inertia changes when the axis of rotation is changed'. Discuss |
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Answer» Solution : MOMENT of inertia of a body I `=MR^(2)=Sigmamr^(2)` `therefore=m_(1)r_(1)^(2)+……….+m_(n)r_(n)^(2)` where is the DISTANCE of each particle of the body from axis of rotation. When axis of rotation CHANGES then distance of the particles will change. So from above EQUATION moment of inertia of the body changes. .Ex : Moment of inertia of thin rod about a perpen-dicular axis passing through centre is `(Ml^(2))/(12)` whereas that passing through one end is `(Ml^(2))/(3)` |
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