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1251.

Lifting a small stone to a certain height and then dropping it downwards. What do you observe?

Answer»


ANSWER :The STONE FALLS down
Discussion
1252.

The weight of the body immersed in a liquid appears to be .............. than its actual weight.

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ANSWER :LESS
Discussion
1253.

A sonar deivce attached to a ship sends. Ultrasomic waves in the sea. These waves are reflected from the bottom of the sea. If the ultrasonic waves take 4 seconds to travel from the ship to the bottom of the sea and back to the ship (in the form of an echol), what is the depth of the sea? (Speed of sound in water = 1500 m//s).

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In a HYPOTHETICAL system , a partical of mass `m` and charge `-3 q` is moving around a very heavy partical chaRGE `q`. Assume that Bohr's model is applicable to this system , then velocuity of mass `m` in the first orbit is `(3 q^(2))/(2 epsilon_(0) h)` `(3 q^(2))/(4 epsilon_(0) h)` `(3 q)/(2 pi epsilon_(0) h)` `(3 q^(2))/(4 pi epsilon_(0) h)` A `(mv^(2))/( r) = (3 q^(3))/(4 pi epsilon_(0) r^(2)) implies m v r = (3 q^(2))/(4 pi epsilon_(0) v)` (i)
and `(nh)/(2 pi) = m v r` (ii)
Using Bohr's (i) and (ii) and putting `n = 1`
`(h)/(2 pi) = (3 q^(2))/(4 pi epsilon_(0) v) implies v = (3 q^(2))/(2 pi epsilon_(0) h)`The binding energy of nucei X and `Y` are `E_(1)` and `E_(2)`, respectively. Two atoms of X fuse to give one atom of `Y` and an energy `Q` is released. Then, `Q=2E_(1) -E_(2)` `Q=2E_(2)-2E_(1)` `Qlt2E_(1) -E_(2)` `Q gt2E_(2) -2E_(1)` b During fusion, binding energy of daughter nucleus is always greater than the total binding energy of the parent nuclei. The difference of binding energied is released. Hence,
`Q=E_(2) -2E_(1)`.A boy is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of focal length `10cm`. The diameter of the sun is `1.39xx10^(9)m` and its mean distance from the earth is `1.5xx10^(11)m`. What is the diameter of the sun's image on the paper ? `9.2xx10^(-4)m` `6.5xx10^(-4)m` `6.5xx10^(-5)m` `12.4xx10^(-4)m` A From the relation `(I)/(O)=(v)/(u)`
Here, `O=1.39xx10^(9),v=0.1m`,
`u=1.5xx10^(11)m`
`:. I=(0.1)/(1.5xx10^(11))xx1.39xx10^(9)=9.2xx10^(-4)m`The temperature of an open room of volume `30 m^(3)` increases from `17^(@)C to 27^(@)C` due to sunshine. The atmospheric pressure in the room remains `1 xx 10^(5) Pa`. If `n_(i) and n_(f)` are the number of molecules in the room before and after heating then `n_(f)` and `n_(i)` will be `2.5 xx 10^(25)` `-2.5 xx 10^(25)` `-1.61 xx 10^(23)` `1.38 xx 10^(23)` B Here, `V_(1) = 30m^(3), T_(1) = 17+273 = 290 K`
`P_(1) = 1 xx 10^(5)Pa`
`V_(2) = 30 m^(3) , T_(2) = 27+273 = 300K`,
`P_(2) = 1 xx 10^(5) Pa`.
Let `N_(1),N_(2)` be the no, of moles of a gas at temperature `T_(1) and T_(2)` respectively. then
`N_(1) =(P_(1)V_(1))/(RT_(1)) = ((1 xx 10^(5)) xx 30)/(83 xx 290) = 1.24 xx 10^(3)`
or `N_(2) = (P_(2)V_(2))/(RT_(2)) = ((1 xx 10^(5)) xx 30)/(83 xx 300) = 1.20 xx 10^(3)` ltbr. Change in the number of moles
`N_(2)-N_(1) =(1.20-1.24) xx 10^(3) = -0.04 xx 10^(3)`
Change in the number of molecules
`n_(f)-n_(i) = (N_(2)-N_(1)) xx (6.023 xx 10^(23))`
`= - (0.04 xx 10^(3)) xx (6.023 xx 10^(23))`
`=-2.5 xx 10^(25)`.An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the Fig.) Initially, the gas is at temperature `T_(1), pressure P_(1) and volume V_(1)` and the spring is in its relaxed state. The gas is then heated very slowly to `temperature T_(2), pressure P_(2) and volume V_(2)`. during this process, the piston moves out by a distance x. ignoring the friction between the piston and the cylinder, the correct statement(s) is (are)
If `V_(2) = 2V_(1) and T_(2)=3T_(1)`, then the energy stored in the string is `1/4 P_(1)V_(1)` If `V_(2) and 2V_(1)` and `T_(2) = 3T_(1)`, then the charge in iternal energy is `3P_(1)V_(1)`. If `V_(2)=3V_(1)` and `T_(2) = 4T_(1)`, then the work done by the gas is `7/3 P_(1)V_(1)` If `V_(2) = 3V_(1)` and `T_(2) = 4T_(1)`, then the heat supplied to the gas is `17/6 P_(1)V_(1)`. B From gas equation , `(P_(1)V_(1))/(T_1) = (P_(2)V_(2))/(T_2) = nR or P_(2) = (P_(1)V_(1)T_(2))/(V_(2)T_(1)) ` ...(i)
(a) when `V_(2) = 2 V_(1), T_(2) = 3T_(1)`, then
`P_(2) = (P_(1)V_(1) xx 3T_(1))/(2 V_(1) xx T_(1)) = 3/2 P_(1)`
If A is the area of cross- section of the piston, then
`P_(2)A = kx or P_(2)Ax = kx^(2)`
Energy in spring , `U = 1/2 kx^(2) = 1/2 P_(2) Ax`
`=1/2 P_(2)(V_(2)-V_(1))`
`=1/2 xx (3P_1)/(2) (2V_(1)-V_(1)) = 3/4 P_(1)V_(1)`
Thus, option (a) is wrong.
(b) When `V_(2) = 2V_(1) and T_(2) = 3T_(1) , P_(2) = 3/2P_(1)` [from (i) ]
Change in internal energy ,
`dU = nC_(v) dT = m((3R)/(2)) (T_(2)-T_(1))`
`= n xx (3R)/(2) [3T_(1)-T_(1)] = 3nRT_(1) = 3P_(1)V_(1)`
Thus option (b) is true.
(c) when `V_(2) = 3V_(1) and T_(2) = 4T_(1)`
`P_(2) = (P_(1)V_(1)xx4T_(1))/(3V_(1)xxT_(1)) = 4/2 P_(1)`
Workdone , dW = energy in spring
`1/2kx^(2) = 1/2 P_(2)(V_(2)-V_(1))`
`1/2 xx 4/3 P_(1)[3V_(1)-C_(1)] = 4/3 P_(1)V_(1)`
Thus option (c) is wrong.
(d) when `V_(2) = 3V_(1) and T_(2) = 4T_(1), P_(2) = 4/3 P_(1)`
Heat supplied to the gas, `dQ = dU + dW`
`dQ = n xx (3R)/(2)[T_(2)-T_(1)]+1/2 P_(2)(V_(2)-V_(1))`
`=3/2 nR [4T_(1)-T_(1)] + 1/2 4/3 P_(1)[3V_(1)-V_(1)]`
`= 9/2 nRT_(1)+4/2 P_(1)V_(1) = 9/2 P_(1)V_(1)+ 4/2 P_(1)V_(1) = 35/6 P_(1)V_(1)`.
Thus, option (d) is wrong.Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at `T_(0)`, while box B contains one mole of He at `7//3 T_(0)`. The boxes are then put into thermal contact with eachother, and heat flows between them till the gases reach a common final temperature `T_(f)`. ignore the heat capacity of boxes. then `T_(f) "in terms of" T_(0)` is `T_(f) = (7)/(3)T_(0)` `T_(f) = (3)/(2)T_(0)` `T_(f) = (5)/(2)T_(0)` `T_(f) = (3)/(7)T_(0)` B Here, `n_(1) "mole" , n_(2) = 1 "mole"`
For nitrogen, `C_(p_1)=7/2 R`
for helium, `C_(p_2) = 5/2 R`
`T_(1)=T_(0), T_(2) = 7/3 T_(0).T_(f)=?`
when gases are put into thermal contact, heat is exchanged between them till final temperature `T_(f)` is reached.
Heat gained by nitrogen = Heat lost by He
`n_(1)C_(P_1)(T_(f)-T_(1)) = n_(2)C_(P_2) (T_(2)-T_(f))`
`(n_(1)C_(P_1)+n_(2)C_(p_2)) T_(f) = n_(2)C_(p_2) T_(2) n_(1)C_(p_1)T_(1)`
`(1 xx 7.2 R + 5/2 R) T_(f) = 1 xx 5/2 R xx 7/3 T_(0) + 1 xx 7/2 RT_(0)`
`6RT_(f) = (35/6 + 7/2)RT_(0)`
`T_(f) = (56)/(6 xx 6) T_(0)~~3/2 T_(0)`.If by mistake Ammeter is connected parallel to the resistance then `i-V` curve expected is (Here `i`= reading of ammeter, `V=` reading of voltmeter)
C As ammeter has very low resistance most of current will pass through the ammeter so reading of ammeter (`i`) will be very large. Voltmeter has very high resistance so reading of voltmeter will be very low.Unit of `(CV)/(rho epsilon_(0))` are of
(`C =` capacitance, `V =` potential, `rho =` specfic resistence and `epsilon_(0) =` permittivity of free space) Charge current time frequency B `C = (epsilon_(0)A)/(d)` and `rho = (RA)/(l)`
Substituting in `(CV)/(rhoepsilon_(0))` we get
`(CV)/(rhoepsilon_(0))= (((epsilon_(0)A)/(d))(V))/((R (A)/(l)) epsilon_(0))-="current"`.Two separate monochromatic light beams `A` and `B` of the same intensity (energy per unit area per unit time) are falling normally on a unit area of a metallic surface. Their wavelength are `lambda_(A)` and `lambda_(B)` respectively. Assuming that all the the incident light is used in ejecting the photoelectrons, the ratio of the number of photoelectrons from beam `A` to that from `B` is `(lambda_(A)/lambda_(B))` `(lambda_(B)/lambda_(A))` `(lambda_(A)/lambda_(B))^(2)` `(lambda_(B)/lambda_(A))^(2)` A The number of photo electron depends on the Number of photons Number of photon `=I/(hc//lambda)=(lambda.I)/(hc) prop lambda`
Ratio of no. of photo electrons =`lambda_(A)/lambda_(B)`The de Broglie wavelength of an neutron corresponding to root mean square speed at `927^(@)` is `lambda`. What will be the de Broglie wavelength of the neutron corresponding to root mean square speed at `27^(@)C`? `(lambda)/(2)` `lambda` `2 lambda` `4 lambda` C `K.E.` of neutron `E=(3)/(2)KT`
`lambda_(d)=(h)/(p)=(h)/(sqrt(2mE))=(h)/(sqrt(2mxx(3)/(2)kT))`
`rArr lambda_(2)=lambdasqrt(((927+273))/(27+273))= 2 lambda`A uniform metal rod is used as a bar pendulum. If the room temperature rises by `10^(@)C` , and the coefficient of linear expansion of the metal of the rod is `2xx10^(-6)` per `.^(@)C` , the period of the pendulum will have percentage increase of `-2xx10^(-3)` `-1xx10^(-3)` `2xx10^(-3)` `1xx10^(-3)` DIf on heating throught `80^(@)C` , the mass expelled is `(1//100)^(th)` of mass still remaining, the coefficient of apparent expansion of liquid is `1.25xx10^(-4)//^(@)C` `12.5xx10^(-4)//^(@)C` `1.25xx10^(-5)//^(@)C` None of these ATwo identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same volume V . The mass of the gas in A is `m_(A)`and that in B is `m_(B).` The gas in each cylinder is now allowed to expand isothermally to the same final volume 2 V . The changes in the pressure in A and B are found to be `DeltaP`and 1.5 `DeltaP` respectively. Then `4m_(A)=9m_(B)` `2m_(A)=3m_(B)` `3m_(A)=2m_(B)` `9m_(A)=3m_(B)` CFollowing figure shows on adiabatic cylindrical container of volume `V_(0)` divided by an adiabatic smooth piston (area of cross-section = A ) in two equal parts. An ideal gas `(C_(p)//C_(y)=lambda)` is at pressure P and temperature T in left part and gas at pressure P and temperature T in right part. The piston is slowly displaced and released at a position where it can stay in equilibrium. The final pressure of the two parts will be (SUPPOSE x = displacement of the piston)
`P_(2)` `P_(1)` `(P_(1)(V_(0)/(2))^(lambda))/((V_(0)/(2)+Ax)^(lambda))` `(P_(2)(V_(0)/(2))^(lambda))/((V_(0)/(2)+Ax)^(lambda))` CIn the FOLLOWING P-V diagram two adiabatics cut two isothermals at temperatures T and T (fig).The value of `(V_(a))/(V_(d))` will be
`(V_(b))/(V_(c))` `(V_(c))/(V_(b))` `(V_(d))/(V_(a))` `V_(b)V_(e)` AA light ray from air is incident (as shown in figure) at one end of a glass fiber (refractive index `mu = 1.5`) making an incidence angle of `60^(@)` on the lateral surface, so that it undergoes a total internal reflection. How much time would it TAKE to traverse the straight fiber of length 1 km
`3.33 mu sec` `6.67 mu sec` `5.77 mu sec` `3.85 mu sec` D N/AA combination of two thin lenses with focal lengths `f_(1)` and `f_(2)` respectively forms an image of distant object at distance 60 cm when lenses are in contact. The position of this image shifts by 30 cm towards the combination when two lenses are separated by 10 cm . The corresponding values of `f_(1)` and `f_(2)` are 30 cm, -60 cm 20 cm, -30 cm 15 cm, -20 cm 12 cm, -15 cm B N/AAn optical instrument uses a 25 D objective and 20 D eyepeice with a tube length of 25 cm when eye is least strained The instrument is a telescope with angular magnification 20. The instrument is a microscope with angular magnification 20. The instrument is a telescope with angular magnification 24. The instrument is a microscope with angular magnification 24. B N/AThe ratio of d-Brogile wavelength off mlecules of hydrogen and helium gas moving at arms speed in two gas jars kept sparately at temeperature `37^(@)` C and `127^(@)` C respectively is : `(2)/sqrt(3)` `sqrt(2/(3))` `sqrt(3)/(2)` `sqrt(8/(3)` DAt t=O, light of intensity `10^(12)` photons/` s-m^(2)` of energy 6eV per photon starts falling on a plate with work founction `2.5 eV` If area of the plate is `2xx10^(-4) m^(2)` and for every `10^(5)` photons one photoelectron is emitted, charge on the plate at t=25 s is `8xx10^(-15) C` `4xx10^(-14) C` `12xx10^(-10) C` `16xx10^(-12) C` ARadius of an electron moving in a circle in constant magnetic field is two times that of an `alpha` particle in the same field. Then de-Broglie wavelength of electrons is x-times of the `alpha`-particle Here x is1A telescope has an objective lens of `10cm` diameter and is situated at a distance of one kilometre from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is `5000 Ã…`, of the order of mm is :5 Resolving limit of telescope is
`theta prop (x)/(D)=(lambda)/(d)implies x=(lambdaD)/(d)`
Given, `lambda=5000 Ã…=5000xx10^(-1)m`
`D=1km=1000m`
`d=10cm=0.1m`
Hence, `x=(5000xx10^(-10)xx1000)/(0.1)`
`=5xx10^(-3)m=5mm`A spring is placed between the jaws of screw gauge such that the spring is not all compressed. The mai scale reads 2 division and circular scale reads 28 divisions. Now we turn the circular scale by `18^(@)` such that the spring is compessed. The circular scale has 200 divisions and the least count of the main scale is `1mm`. the force exerted by the spring on the jaws in unit `mN`. (the spring constnat is `100N//m`)5Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects than human eye can resolve at 500nm wavelength is `X mm` , so value of `x` is:30
`sintheta=(0.25)/(25)=(1)/(100)`
Resolving power `=(1.22lamda)/(2musintheta)=30mum`In a cyclic process shown in the figure an ideal gas is adiabatically taken from `B` and `A`, the work done on the gas during the process `B rarr A` is `30 J`, when the gas is taken from `A rarr B` the heat absorbed by the gas is `20 J`. The change in internal energy of the gas in the process `A rarr B` is given as `-X`, so value of `X` is - :
30 N//A

Solution :The time taken by the ultrasonic sound waves to travel from the ship to the sea-bed, and back to the ship is 4 seconds. So, the time taken by the ultrasonic sound to travel from the ship to sea-bed will be half of this time, which is `(4)/(2) =2` second. This means that the sound takes 2 seconds to travel from the ship to the bottom of the sea.
Now, Speed `= ("Distance")/("Time")`
So, `1500 = ("Distance")/(2)`
And, Distance `= 1500 xx 2m = 3000m`
Thus, the depth of this sea below the ship is 3000 METRES.
Discussion
1254.

State two applications of universal law of gravitation.

Answer»
Discussion
1255.

What is the speed of the body moving with uniform acceleration at the midpoint of two points on a straight line, where the speeds are u and v respectively?

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Solution :LET .a. be the CONSTANT acceleration and s be the distance between the two points, From equation of motion
`V^(2) - u^(2) = `2as ............ (1)
Let `v_(0)` be the speed of the body at midpoint ‘M’ of the given points. Applying the same equation used above, we get
`v_(0)^(2) - u^(20 = 2a (s)/(2)`
From (1) , we get
`v_(0)^(2) - u^(2) = (v^(2) - u^(2))/(2)`
`v_(0)^(2) = (v^(2) - u^(2))/(2) + u^(2)`
`v_(0)^(2) = (v^(2) - u^(2) + 2u^(2))/(2)`
` v_(0) = sqrt((v^(2) + u^(2))/(2))`
Discussion
1256.

Under what condition (s) is the magnitude of average velocity of an objectis equal to its average speed ?

Answer»

Solution :If an OBJECT moves in a straight LINE in the same DIRECTION, then the magnitude of average velocity of an object is equal to its average SPEED.
Discussion
1257.

A scooter or a motor-cycle is a compoundmachine made up of several simple machines. Study the following parts of a scooter/motor cycle and identify the simple machines involved in them. (i) Clutch levers(ii) Throttle (iii) Front brake lever(iv) Rear brake pedal (v) Stand(vi) Rear view mirror (vii) Wheels(viii) Chain drive A person weighing 50 kg, moves on a scooter of 100 kg at a speed of 36 km h^(-1) and applies brakes to stop within a distance of 10m. If the mechanicaladvantage of the brake system (comprising brake drum, lever, etc.) is 10^(3), find the force with which the person should press the foot pedal to stop the vehicle.

Answer»

SOLUTION :(i) Equations of motion, Newton's SECOND law of motion and principleof moments.
(ii) Calculate retardation using` a = (V^(2)-u^(2))/(2S)`
`(iii) Calculate the braking (retarding ) force using F = M.A., where 'm' is the combined mass of the SCOOTER and the person.
(iv) Calculate effort using:
effort = load/M.A. = F/M.A.
(v) 1.5 N
Discussion
1258.

Compare the time periods of two simple pendulums of length 1 m and 16 m at at place.

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Solution :Given:`l_1 = 1 m and l_2 = 16 m`
Since `T PROP sqrtl THEREFORE (T_1)/(T_2) = sqrt((l_1)/(l_2))`
`therefore (T_1)/(T_2) = sqrt(1/(16)) = 1/4 `i.e. `T_1: T_2= 1:4`
Discussion
1259.

Find the energy in kwh consumed in 10 hours by four devices of power 500w each.

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Solution :ENERGY consumed by an electric DEVICE can be obtained with the help of the expression for power,
P = w/t
Where power rating of the device, p = 500w = 0.50 kw
Time for which the device runs, T = 10h. Work done =energy consumed by the device.
Therefore, energy consumed = Power `xx` Time = `0.50 xx 10 = 5kwh`
Hence, the energy consumed by FOUR equal rating devices in 10h will be `4 xx 5 KWH = 20 kwh - 20 units`
Discussion
1260.

When a stone of mass 50 kg and another of mass 5kg fall down simultaneously form the top of a fivestorey building, which one will reach the ground first?

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ANSWER :They reaches the GROUND at the same TIME.
REASON: Acceleration due to gravity does not depend on the mass of the body. It will be the same for allGMbodies falling to the earth. `g=GM/R^2`
Discussion
1261.

Which of the following has more inertia:

Answer»

RUBBER BAIL
onion
a five-rupees COIN
a ROCK

ANSWER :d
Discussion
1262.

A train starting from a railway station attains a velocity of 30 ms^(-1) in one minute. What is its acceleration ?

Answer»

SOLUTION :`a = (v-u)/(t)`
Here,` u =0, v = 30 MS ^(-1) and t= 1 min = 60 sec = (30-0)/(6) = 0.5 ms ^(-2).`
The acceleration PRODUCED in the train 3 is 0.5 `ms^(-2).`
Discussion
1263.

A bus travels, a distance of 20 km from Chennai central to airport in 45 minutes. What is the average speed?

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Solution :Given : Distance =20 km =20,000 m
Time =45 min=2700 s
AVERAGE SPEED =`"Total Distance"/"Total Time Taken"`
Average speed = `"20 km"/"40 min"="20,000m"/"2700 s"`
`="200 m"/"27 s"=7.4 ms^(-1)`
Discussion
1264.

An artificial satellite is moving in a circular orbit of radius 42,250 km. Calculate its speed if it takes 24 hour to revolve around the earth.

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Solution :Here,` r = 42,250 KM = 42250 XX 1000 m `
Time taken for one rovolution
`= 24 H = 24 xx 60 xx 60 s`
Speed (u) `= ("Distance")/("Time") = (2pi r )/("Time") = (2 xx (22)/(7)xx 42250 xx 1000 m )/( 24 xx 60 xx 60 s) `
`~~3073.74 ms ^(-1)`
The speed of the satellite is `3073.73 ms ^(-1)`
Discussion
1265.

A boy holds a convexlens 30 cm above the base of an empty vessel. Thereal image of thebottom of thevessel is formed20 cm abovethe lens.The boy fills aliquid in thevessel up to a depth of 25 cm and finds thatthe real image of thebottom of the vessel is now 30 cm above the lens. Find the refractive index of theliquid.

Answer»

Solution :(i) Let the focal length of the lens be f.
` f = (v_(1)v_(1))/(v_(1) - u_(1)) `
Determine the value of f andlet THEBOTTOM of the vessel appear to be at a DEPTH `'u_(2)'` from thelens when the beaker is filled with a liquid.
` 1/u_(2) = 1/v_(2) - 1/f`
where `v_(2)` is theimage distance when thebeaker is filled with water.
Now theshift of thebottom of the vessel= `u_(1) - u_(2)`
The apparent depth of thewater in the container= `d - (u_(1) - u_(2))` where 'd' is thedepth of thewater in thecontainer.
`:. " REFRACTIVE index " = ("Real depth")/("Apperent depth") = d/(d-(u_(1)-u_(2)))`
(ii) ` mu = 1.66or 5/3`
Discussion
1266.

Fill in the following blanks with suitable words : (a )Force acting on a units area is called …….. (b) It is the …..force which makes objects appear lighter in water (c ) A heavy ship floats in water because its ………….density is less thanthat of water . (d ) In fluids (liquids and gases ), pressure acts in .........direction ,and pressure .....as the depth increases, (e ) In order to sink in fluid ,the density of an object must be .........than the ...... of the fluid . (f ) snow shoes work by spreding out a preson's ........over a much bigger ..........

Answer»


Answer :(a) PRESSURE (b )buoyant (c ) AVERAGE(d) all, increase (E ) more ;dencity ( F) weight area
Discussion
1267.

A car moves with constant speed of 10 m/s in a circular path of radius 10m. The mass of the car is 1000 kg. How much is the required centripetal force for the car?

Answer»

SOLUTION :`10^(4)N`
Discussion
1268.

The distance between Ahmedabad and Vadodara is 100 km. A scooterist travels at the average speed of 50 km h^(-1). If he travels first 60 km at a uniform speed of 45 kmh^(-1), what will be his speed for the remaining distance ?

Answer»

SOLUTION :`60 KM H ^(-1)`
Discussion
1269.

Calculate the weight of a body in the moon if it weights 150 kg in earth?

Answer»


ANSWER :WEIGHT in the MOON = `150xx1/6`= 25 KG
Discussion
1270.

Which of the following can produce longitudinal waves as well as transverse waves under different conditions?

Answer»

water
TV transmitter
slinky
tuning fork

Answer :C
Discussion
1271.

Name the mirror(s) that can give (I) am erect and enlarged image, (ii) same sized, inverted image.

Answer»

SOLUTION :CONCAVE MIRROR
Discussion
1272.

State Newton's laws of motion

Answer»

Solution :I law : An object remain in a stateof rest or a uniform motion in a straight line unless ACTED UPON by an external unbalanced force.
II law : The rate of momentum of an object is proportional to the APPLIED unbalanced force in the direction of the force.
III law : To every action , there is an equal and OPPOSITE reaction and they act on two DIFFERENT bodies.
Discussion
1273.

Which of the following device does not work on reflections of sound waves?

Answer»

stethoscope
hydrophone
soundboard
megaphone

Answer :B
Discussion
1274.

Draw a graphical represntation of the wave shape for (a) low pitched sound (b) a high pitched sound.

Answer»



SOLUTION :
Discussion
1275.

How do green plants produce food?

Answer»

Solution :The green leaves of plants USE solar ENERGY to MAKE their food through the PROCESS of photosynthesis.
Discussion
1276.

When an object is fully submerged in strong salty water, it undergoes an apparent:

Answer»

LOSS in mass
loss in volume
loss in density
loss in WEIGHT

ANSWER :D
Discussion
1277.

What are electric lines of force?

Answer»

Solution :The direction of the electric FIELD is the direction OFTHE force that WOULD ACT on a small positive charge. Therefore, the lines REPRESENTING the electric field are called .electric lines of force..
Discussion
1278.

Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room ?

Answer»

SOLUTION :QUALITY (or TIMBRE) of SOUND.
Discussion
1279.

The mass per unit volume of a substance is called density.

Answer»

SOLUTION :(density = mass / VOLUME)
Arrangethe following in order of increasing density.
Air, EXHAUST from chimney, HONEY, water, chalk, cotton and iron.
Exhaust from chimney
`{:("Air", "Water"),("Honey", "Chalk"),("Cotton", "Iron"):}`
Discussion
1280.

What are the effects of electric current?

Answer»

Solution :when current FLOWS in a circuit, it exhibits various effects. The MAIN effects are, HEATING, chemical and MAGNETIC effects.
Discussion
1281.

An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s-1 to 8 m s ‘ in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Answer»
Discussion
1282.

At what speed a person having 60 kg mass should run to acquire 750 J kinetic energy?

Answer»


ANSWER :`5MS^(-1)`
Discussion
1283.

A motorcycle travelling at 20 ms^(-1) has an acceleration of 4 ms^(-2). What does it explains about the velocity of the motorcycle?

Answer»

SOLUTION :It explains that the MOTORCYCLE TRAVELS with uniform velocity.
Discussion
1284.

In an experiment to determine the densities, four solids A,B,C and D are immersed in a liquid contained in a measuring cylinder one by one. The volume of water displaced by the solids A,B, C and D are 100 cm^(3), 100 cm^(3), 80 cm^(3) and 80 cm^(3), respectively. When weighed in air, the masses of solids A,B,C and D were found to be 80g,100g,100g and 80g, respectively. The two solids having identical densities are:

Answer»

A and C
B and D
C and D
A and D

Answer :B
Discussion
1285.

If 15 g of a saturated solution of sodium chlroide at 20^(@)C, leaaves a solid residue of 3 g, when evaporated to dryness, calculate the solubility of NaCl at that temperature.

Answer»

Solution :WT. of water in solution=`15-3=13g,""So," "12g" "H_(2)O` DISSOLVES 3g of SOLID
`therefore 100" g "H_(2)O" dissolves "(3)/(12)xx100g` of solid `=25g``therefore` solubility in % (w/w)=25.
Discussion
1286.

Does sound follow the same laws of reflection as light does ? Explain.

Answer»



Solution :Sound follows the same laws of REFLECTION ASLIGHT does. The incident sound wave and the reflected sound wave MAKE the same angle with the normal to the surface at the point of incidence. ALSO , the incident sound wave, the reflected sound wave, and the normal to the point of incidence all LIE in the same plane.
Discussion
1287.

In a 100 m race, the winner takes 10s to reach the finishing point. The average speed of the winner is ______ ms^(-1)

Answer»

5
10
20
40

Answer :B
Discussion
1288.

An object travels 16 m in 4s and then another 16 m in 2 s. What is the average speed of the object?

Answer»

Solution :Total distance travelled by the object
`= 16 m + 16 m = 32 m`
Total time taken `= 4 s + 2S = 6s`
Average speed `V _(av) = ("Total distance travelled")/("Total time taken") = (32m)/(6s) = 5.33 MS ^(-1)`
`THEREFORE` Therefore, the average speed of the object is `5.33 ms ^(-1).`
Discussion
1289.

Speed (s), wave length(lambda) and frequency (upsilon)of sound are related as

Answer»

`S= U XX UPSILON`
`v=s xx u`
`u = s xx upsilon`
`u=s//upsilon`

ANSWER :C
Discussion
1290.

A man first swims in sea water and then in river water. (i) compare the weights of sea water and river water displaced by him. (ii) Where does he find it easier to swim and why?

Answer»


Answer :(i) 1:1 (ii) In sea WATER because the density of sea water is more than that of RIVER water so his WEIGHT is balanced in sea water with his less submerged INSIDE it.
Discussion
1291.

Rubbing a comb on hair makes the comb get -0.4C. (a) Find which material has lost electron and which one gained it. (b) Find how many electrons are transferred in this process.

Answer»

Solution :(a) The material is COMB which GAINED electrons and the HAIR gained it.
(b) Charge on 1 electron, `e= 1.6xx10^(-19)C`
`q= "ne or "n=q"/"e`
`n= 0.4"/"1.6xx10^(-19)=0.25xx10^(19)`
`=2.5xx10^(18)`.
Discussion
1292.

What is a positive and a negative ion?

Answer»

Solution :If an ELECTRON is removed from the atom, it BECOMES POSITIVELY charged. Hence it is CALLED a positive ion. When an electron is added to an atom, it becomes negatively charged. Hence it is called a negative ion.
Discussion
1293.

Find out the displacement between 10 s and 20 s from the graph.

Answer»


ANSWER :DISPLACEMENT = AREA of ABCD=`ABtimesAD`=`20times10`=200 m
Discussion
1294.

A hydraulic system is used to lift a 2000 kg vehicle in an auto garage. If the vehicle sits on a piston of area 0.5 m^(2) ,m, and a force is applied to a piston of area 0.03m^(2) , what is the minimum force that must be applied to lift the vehicle?Given: Area covered by the vehicle on the piston A_(1) = 0.5 m^(2) Weight of the vehicle , F_(1) = 2000 kg xx 9.8 m s^(-2) Area on which force F_(2) is aplied , A_(2) = 0.03 m^(2)

Answer»

Solution :`P_(1) = P_(2) , F_(1)/A_(1) = F_(2)/A_(2) " and " F_(2) = F_(1)/A_(1) A_(2) ` ,
` F_(2) = (2000 XX 9.8) (0.03)/(0.5) = 1176 ` N
Discussion
1295.

A rectangle glass wedge (prism) is immersed in water as shown in figure E-a. For what value of angle alpha, will the beam of light, which is normally incident on AB, reach AC entirely as shown in figure E-b. Take the refractive index of water as 4/3 and the refractive index of glass as 3/2.

Answer»

Solution :From the geometry of figure E-b it is CLEAR that, the angle of incidence on the side BC is EQUAL to `alpha` (dotted line is a normal drawn at the POINT of incidence). The ray should undergo total internal reflection to reach AC. For occurrence of total internal reflection, the value of `alpha` must be greater than the critical angle at interface of water and glass.
LET .C. be the critical angle of interface of glass and water. From the given condition `alpha gt C .......(1)`
We know, `sin C=1/(n^(12))` ........(2)
`n_(12)=3/2 / 4/3=9/8`
From equation 2, we get
`sin C=8/9 rArr C=62^(@)30^(1)`
Hence `alpha` is greater than `C=62^(@)30^(1)`
Let us see few activities of total internal reflection.
Discussion
1296.

A motorcycle moving with a speed of 5 ms^(-1)is subjected to an acceleration of 0.2 ms^(-2) for 10 second. Calculate (a) the speed (b) the distance travelled.

Answer»

Solution :Here,` u = 5 ms ^(-1),a =0.2 ms ^(-2), t = 10` second
`(a) v = u + at `
`= 5 ms ^(-1) + 0.2 ms ^(-2) xx 10 s `
` = 5 ms ^(-1) + 5 ms ^(-1) = 10 ms ^(-1)`
(b)` s = ut + 1/2 at `
`= 5 ms ^(-1) + 10S + 1/2 xx 0.2 ms ^(-2) xx (10 s ) ^(2)`
`= 50 m + 10 m = 60 m`
The speed of the MOTORCYCLE after 10 second is `10 ms ^(-1),` the distance TRAVELLED by the motorcycle is 60 m.
Discussion
1297.

Suppose the weight of a man on the planet A is half than that on the earth. If he can jump 0.4 m high on earth's surface, then how high he can jump on the planet A?

Answer»


Solution :The gravitational POTENTIAL energy of the man is same on the planet A and earth.
`therefore (mg_(1)h_(1))_("earth")=(mg_(2)h_(2))_("planet "A)`
`therefore (mg_(1))h_(1)=((mg_(1))/(2))h_(2)`
`therefore 2h_(1)=h_(2)`
`=2xx0.4m`
`therefore h_(2)=0.8m`
Discussion
1298.

When the reflecting surface is curved outwards the mirror formed will be ....................... ·

Answer»

CONCAVE mirror
Plane mirror
Convex mirror

Answer :B
Discussion
1299.

State the units of speed

Answer»

Solution :The SI units of SPEED is m/s `(ms^(-1)).` The unit km/H (`km h^(-1))` is also USED.
Discussion
1300.

A person walks 50 m in the north direction and from there he walks 30 m in the west direction. Now, he/she walks 50 m distance in south direction. Find the distance travelled by the person and his/her displacement. Find out the average speed and average velocity of person if he/she takes 100 s time to move from initial to final position.

Answer»

Solution :Total DISTANCE = 130m, displacement =0
Average SPEED `= 1.3 ms ^(-1),`
Average velocity `= 0.3 ms ^(-1)`
Discussion