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1501.

A sound wave travels at a speed of339ms^(-1) . If its wave length is 1.5cm, What is the frequency of the wave ? Will it be audible?

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Solution :Speed of sound, `upsilon=3.39ms^(-1)`
WAVE length of sound, `lambda=1.5cm=0.015m`
Speed of sound = wave length x FREQUENCY `= lambda xx upsilon`
Therefore `upsilon=upsilon//lambda = 339//0.015=22600` Hz.
The frequency range of AUDIBLE sound for humans LIES between 20Hz to 20,000 Hz. Since the frequency of the given sound is more than 20,000 Hz, it is not audible.
Discussion
1502.

______ is the value of universal gravitational constant.

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SOLUTION :`G=6.7xx10^(-11)NM^(2)KG^(-2)`
Discussion
1503.

The wavelength of light is 5000 Å , express it in m .

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SOLUTION :`5XX10^(-7) m`
Discussion
1504.

Avernierscalewith 10divisions slidesovera mainscalewhosepitchis 0.5 mm (pitch =1M.S.D) . If abob ofdiameter 9.75 mm is held betweenthhe jaws determinethe MSRand V.S.D . If (a)there is no zeroerror (b) the zeroerror= 0.35 mm

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Solution :Pitch `=(5 mm)/(10)`= 0.5mm, LEAST count
`=(0.5)/(N) =(1)/(2N)`
Diameter =PSR +C.S.C.D. `xx` L.C.
`=xp+ y.(P)/(N)= P(x+.(y)/(N))`
`x+ .(y)/(N) =5.48`
SINCE `.(y)/(N) lt ` pitch `x+ .(y)/(N)=5+0.48` ltbgt x=5
`(y)/(N)=0.48 rArry=0.48 N`
Thusthe CSCDdependson thenumber THEOF divisions on thecircular scale forexampleif thenumberof circular scaledivisionsis 100the CSCDis 48 P.S.R=5
`L.C. =(1)/(2N)`
`C.S.C.D = 0.48 N`
Discussion
1505.

A simple pendulum completes 40 oscillations in one minute. Find its frequency

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SOLUTION :`0.67 s^(-1)`
Discussion
1506.

In which of the three media, air, water iron, does sound travel the fastest at a particular temperature ?

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Solution :The speed of SOUND depends on the nature of the medium sound travels the FASTEST in solids. It speed decreases in LIQUIDS and it is the SLOWEST in gases. Therefore for a given temperature sound travels faster in IRON.
Discussion
1507.

Sonar makes use of :

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INFRASONIC sound
ULTRASOUND
ORDINARY sound
light

Solution :ultrasound
Discussion
1508.

In a certain gas, a source produces 40,000 compression and 40,000 rarefaction pulses in 1 sec. When the second compression pulse is produced, the first is 1cm away from the source. Calculate the wave speed.

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Solution :We know frequency is equal to number of COMPRESSION or rarefaction pulses travelled per second, hence frequency `(UPSILON)=40,000` HZ
Wave length `(lambda)=` distance between two consecutive compression or rarefaction pulses.
`lambda=1` CM
From `v= upsilon lambda = 40,000 Hz xx 1 cm= 40,000 cm//s=400 m//s`
Discussion
1509.

A box of 20kg mass is pulled by force F with constant velocity on a horizontal surface. If force of friction is 49 N, work done during displacement of 10m is………….J.

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490
245
980
zero

Solution :The BOX moves with constant velocity, i.e., BALANCED forces are acting on it. The resultant force of balanced forces is zero. Hence, work being DONE during displacment of the box WOULD be zero.
Discussion
1510.

Do the velocity , frequency and wavelength of a sound wave increase , decrease or remain constant , when it is reflected from an obstacle ? Explain .

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SOLUTION :VELOCITY of SOUND in a GIVEN MEDIUM
Discussion
1511.

Name the instrument which has the least count 1 mm .

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SOLUTION :METRE RULE
Discussion
1512.

Which characteristicsof the sound helps you to identify your friend by his voice whilesitting with other in a dark room ?

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Solution :The quality or TIMBER SOUND ENABLES us to IDENTIFY our FRIEND by his voice.
Discussion
1513.

Which wave property determines (a) loudness (b) pitch ?

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SOLUTION :(a) AMPLITUDE
(B) FREQUENCY
Discussion
1514.

A wave pulse of frequency 200 Hz, on a string moves a distance 8 m in 0.05 s. Calculatethe velocity of pulse .

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SOLUTION :GIVEN, d = 8 m , t = 0.05s, f= 200 Hz
The velocity of PULSE V =` ("Distance moved d")/("Time TAKEN t")`
or `V= (8 m)/(0.05s) = 160 m s^(-1)`
Discussion
1515.

Calculate the heat energy required to raise the temperature of 2 kg of water from 10^@C to 50^@C. Specific heat capacity of water is 4200 JKg^(-1) K^(-1)

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Solution :Given m=2kg , `DELTAT`=(50-10)=`40^@C`
In terms of KELVIN, `DeltaT`=(323.15 - 283.15)=40 K, `C=4200 J KG^(-1) K^(-1)`
`therefore` HEAT energy required , `Q=mxxCxxDeltaT=2xx4200xx40`=3,36,000 J
Discussion
1516.

What is common to both direct and alternating current?

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Solution :Joule.s HEATING EFFECT of current is COMMON to both direct and alternating current.
Discussion
1517.

Define: (a) Thrust, (b) Pressure

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Solution :(a) Force ACTING on a BODY perpendicular to the surface is CALLED thrust.
(b) The force per unit AREA acting on an object CONCERNED is called pressure.
Discussion
1518.

What is the concept of temperature?

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SOLUTION :Temperature is the DEGREE of hotness or coolness of a body. The HOTTER the body is HIGHER is its temperature.
Discussion
1519.

Find the energy in kWh consumed in 10 hour by 4 devices of power 500 W each.

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Solution :Here, `p=500W=0.05kW,`
`t=10` hour
ENERGY CONSUMED by FOUR devices
`=4pxxt=4xx0.5kWxx10h`
`=20kWh`
Discussion
1520.

The primary coil of a transformer has 800 turns and the secondary coil has 8 turns. I connected to a 220 V as supply. What will be the output voltage?

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<P>

SOLUTION :In a transformer `E_(s)//E_(p)=N_(s)//N_(p)`
`E_(s)=N_(s)//N_(p)xxE_(p)`
`=8//800xx220=220//100=2.2` VOLT
Discussion
1521.

A man walks 1.5 m in the east direction, then he walks 2.0 m in south direction and at last walks 4.5 m in the east direction, then using appropriate scale (e.g., 1 cm = 1 m) using scale... ( i ) Calculate the distance covered by the man. (ii) Find the resultant displacement (magnitude) of the man.

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SOLUTION :(i) 8M
(II) 6.3 m`
Discussion
1522.

The most non-polluting and efficient lighting device is:

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CFL
LED
Fluorescent light
Electric bulb

Answer :D
Discussion
1523.

State the units in which pressure is measured .

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Discussion
1524.

A car having mass 1000 kg is moving with speed of 36kmh^(-1). Calculate its momentum and kinetic energy. (Verify the value of kinetic energy using E_(k)=p^(2)//2m. 'p' is momentum, how kinetic energy is equal E_(k)=p^(2)//2m? Think)

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Solution :`p=mv and E_(k)=(1)/(2)mv^(2)`, USE these formulae.
`E_(k)=(1)/(2)mv^(2)XX(m)/(m)`
`=(1)/(2)(m^(2)v^(2))/(m)=(p^(2))/(2M)`
Discussion
1525.

The speed-time graph for a car is shown is Figure (a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during this period. (b) Which part of the graph represents uniform motion of the car ?

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Solution :(a) The distance travelled by the CAR inthe first4 seconds is given by the area between the speed-time curve and the time axis from `t` = 0 to `t` = 4 s. This area of the distance-time graph which represents the distance travelled by the car has been shaded in the graph shown on the next page.
In order to find the distance travelled by the car in the first4 seconds, WEHAVE to count the numberof squares in the shaded part of the graph and also calculate the distance represented by one square of the graph paper. While counting the number of squares which are half or more than half are counted as complete squares but the squares which are LESS than half are not counted. When counted in this way, the total number of squares in the shaded part of the graph is FOUND to be 63.

We will now calculate the distance representedby 1 square of the graph. This can be DONE as follows : If we look at the X-axis, we find that 5 squares on X-axis represent a time of 2 seconds.
Now, `""` 5 squares on X-axis = 2 s
So, `""` 1 square on X-axis = `(2)/(5)` s `""` ...(1)
Again,if we look at the Y-axis, we find that 3 squares on Y-axis represent a speed of 2 m `s^(-1)`.
Now, `""` 3 squares on Y-axis = 2 m `s^(-1)`
So, `""` 1 square on Y-axis = `(2)/(3)`m `s^(-1)` `""`...(2)ltBrgt Since 1 square on X-axis represents `(2)/(5)` s and 1 square on Y-axis represents `(2)/(3)" m "s^(-1)`, therefore :
`""` Area of 1 square on graph represents a distance= `(2)/(5)" s"xx(2)/(3)" m "s^(-1)`
`"" =(4)/(15)` m
Now, `""` 1 square represents distance = `(4)/(15)` m
So, `""` 63 squares represent distance = `(4)/(15)xx63` m
Thus, the car travels a distance of 16.8 metres in the first 4 seconds.
(b) In uniform motion, the speed of car becomes constant. The constant speed is represented by a speed-time graph line which is parallel to the time axis. In the given figure,the straight line graph from `t` = 6 s to `t` = 10 s represents the uniform motion of the car. The part of graph representing uniform motion has been labelled AB.
Discussion
1526.

A cell of potential difference 12 V is connected toa bulb. The resistance of filament of bulb when itglows is 24 Omega. Find the current drawn from the cell.

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ANSWER :`0.5 A`
Discussion
1527.

A ball of mass 0.5 kg thrown upwards reaches a maximum height of 5m.Calculate the work done by the force ofgravity during thisverticaldisplacement considering the value of g =10m//s^(2).

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Solution :Force of GRAVITY acting on the ball,
`F = mg = 0.5 xx 10 = 5N`
Displacement of the ball, S = 5 m
The force and displacement are in opposite DIRECTIONS. Hence WORK DONE bythe force of gravity on the ball is negative.
`W = -FS`
` = -5 xx 5`
` = -25J`
Discussion
1528.

How are the following derived units related to the fundamental units ? (a)Newton , (b)Watt , (c ) Joule , (d) Pascal

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SOLUTION :`kgm^2 s^(-2)`
Discussion
1529.

A car travels from rest with a constant acceleration "a" for "t" seconds. What is the average speed of the car for its journey if the car moves along a straight road?

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SOLUTION :The car STARTS from rest, so u = 0 The distance covered in time t
s = `(1)/(2) at^(2) `
AVERAGE speed= `(" TOTAL distance " )/("Time taken " )`
`v = (("at"^(2)))/(t)`
= `(at)/(2) `
Discussion
1530.

A truck starts from rest and rolls down a hill with constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force on it if its mass is 7 metric tonnes.

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Answer :there is no unbalanced force on the ball so that ball WOULD WANT to come to REST.
Discussion
1531.

What are cathode rays? State their specific charge (charge to mass ratio). What is the charge on an electron ?

Answer»
Discussion
1532.

Lifting a small stone to a certain height and then dropping it downwards. What change takes place in the speed of the stone as it is thrown up?

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ANSWER :SPEED DECREASES
Discussion
1533.

Thre identical blocks, each of mass 10kg, are pulled as shown on the horizontal frictionless surface If the tension (F) in the rope is 30N. What is the acceleration of each block? And what are the tension in the other ropes? (Neglect the masses of the ropes) (A S_(1))

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SOLUTION :`a =1m//s^(2), T_(1) =10N, T_(2) =20N)`
Discussion
1534.

Observe the figures of an object placed in different liquids Among A and B,which is the liquid whose density is greater than that of the object? Why? .

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Answer :LIQUID A. As the BODY floats, the density of the liquid is GREATER.
Discussion
1535.

An object experience a net zero exteranal unbalanced force. Is it possible for the object to be travelling with a non zero velocity ? If yes, state the condition that must be placed on the magnitude and direction of the velocity. If no, provide reason.

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Solution :YES, an object may TRAVEL with non-zero VELOCITY even when the NET external force on it is zero. A rain dropfalls down with a constant velocity. The WEIGHT of the drop is balanced by the upthrust and the velocity of air.The net force on the drop is zero.
Discussion
1536.

Distinguish between speed and velocity. (AS_(1))

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SOLUTION :
Discussion
1537.

Earth's geographic north pole is vary pole close to its magnetic ____________

Answer»


ANSWER :SOUTH POLE
Discussion
1538.

The pitch of a screw gauge is 0.5 mm and the head scale is divided in 100 parts. What is the least count of screw gauge ?

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SOLUTION :0.005 MM or 0.0005 CM.
Discussion
1539.

The area under velocity - time graph represents the _____

Answer»

velocity of the MOVING object
DISPLACEMENT COVERED by the moving object
Speed of the moving object
None

Answer :B
Discussion
1540.

The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

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Solution :The potential energy of a freely falling object DECREASES progressively. As the object falls down, its VELOCITY INCREASES. HENCE the kinetic energy of the object increases. The total energy `(E_(p)+E_(K))` remains constant. The law of conservation of energy is not violated.
Discussion
1541.

Find out reasons for the situations Place some carom board coins in a pile. Using the striker, strike out the coin at the bottom. What do you observe ? What is the reason ?

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Answer :Only the coin at the bottom is thrown AWAY. OTHERS will ramain in the PREVIOUS STAGE.
Discussion
1542.

What is sound and how is it produced?

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SOLUTION :Sound is a form of energy which produces the sensation of hearing. It is produced due to vibrations of different OBJECTS, EG. Stretched vibrating WIRES and vibrating drums, etc.
Discussion
1543.

What is the time period of satellite near the earth's surface? (neglect height of the orbit of satellite from the surface of the earth)?

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Solution :The force on the satellite DUE to earth is given by `F=(GmM)/(R^(2))`
M-Mass of earth `=6xx10^(24) kg`
m-mass of satellite,
R-radius of earth `=6.4xx10^(6) m`
LET v be the speed of the satellite
`v=(2pi R)/(T) rArr T=(2 pi R)/(v)`
Required centripetal force is provided to satellite by the gravitational force hence
`F_(C )=(mv^(2))/(R)`
But `F_(C)=(GMm)/(R^(2))` according to Newton.s LAW of gravitation.
i.e., `(GMm)/(R^(2))=(m(2pi R)^(2))/(T^(2)R)`
`rArr T^(2) =(4 pi^(2) R^(3))/(GM)`, as mass of the earth (M) and G are constants the value of T depends only on the radius of the earth.
`rArr T^(2) alpha R^(3)`
SUBSTITUTING the values of M, R and G in above equation we get, T = 84.75 minutes. Thus the satellite revolving around the earth in a circular path near to the earth.s surface takes 1Hour and 24.7 minutes approximately to complete one revolution around earth.
Discussion
1544.

An object of mass 40 kg is raised to a height of 5m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down.

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Solution :Gravitational potential energy is given by the EXPRESSION, w = mgh Where, h = vertical displacement = 5m
mass of the object = 40 kg
g = Acceleration due to GRAVITY = `9.8 ms^(-2)`
Therefore w = `40 xx 5 xx 9.8 = 1960 J`.
At half-way down, the potential energy of the objects will be 1960/2 = 980 J.
At this point, the object has an equal amount of potential and kinetic energy. This is due to the law of CONSERVATION of energy.
Hence, half way down, the kinetic energy of the object will be 980 J.
Discussion
1545.

On pressing a spring its potential energy ………….. (remains constant, increases, decreases)

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ANSWER :INCREASES
Discussion
1546.

A ball is gently dropped from a height of 20m. If its velocity Increases uniformly at the rate of 10 ms^(-2) . With what velocity will it strike the ground? After what time will it strike the ground?

Answer»

Solution :Given : height =20 m
acceleration =`10 MS^(-2)`
Formula : For free falling body, (i) V = gt (ii) s=1/2 `"gt"^2`
time TAKEN to strike the GROUND =`s=1//2 "gt"^2`
`20 m=1//2xx 10ms^(-2) xxt^2`
`r^2="40 m"/(10 ms^(-2))=4s^2`
`therefore` t=2s
(ii)Velocity of the ball when it STRIKES ground v=gt
`v=10 ms^(-2) xx2s`
`v=20ms^(-1)`
Discussion
1547.

An object just floats in water. If common salt is added into the water,

Answer»

the VOLUME of the OBJECT immersed in the liquid decreases.
the object SINKS
the object first sinks and then floats up.
cannot be determined

ANSWER :A
Discussion
1548.

What is crest and trough?

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Solution :When a wave is propogated as REPRESENTED below. A PEAK is CALLED the crest and a valley is called the trough of a wave.
Discussion
1549.

What is the relationship between kinetic energy and work?

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Solution :Work done by the resultant force acting on a moving object is equal to change in kinetic ENERGY of the particle.
i.e, if v is the final velocity and u is the initial velocity then work done (w)
`w = Delta E_(k) = E_(k)` (final VEL) - EK (initial vel) = `1/2 mv^(2) - 1/2 mu^(2)`
it is also called as work energy THEOREM.
Discussion
1550.

Density of mercury is 13600 kg m^(-3) .Calculate the relative density.

Answer»

Solution :Relative DENSITY =Density of Mercury/Density of WATER at `4^(@) C`
` R.D = 13600 //1000`
` = 13.6`
Discussion