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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 13801. |
A particle is displaced from a position (2hati - hatj+hatk) to another position (3hati+2 hatj -2hatk) under the action of the force (2 hati + hatj - hatk). The work done by the force in arbitrary unit is |
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Answer» 8 `vecF = 2 hati + hatj - hatk` Displacement, `vecr = vec(r_2) - vec(r_1) = (3hati + 2hatj - 2hatk) - (2hati - hatj + hatk)` `= hati + 3hatj - 3hatk` `W = vec(F).vec(r) = (2hati + hatj - hatk).(hati + 3hatj - 3hatk) = 2 + 3 + 3 = 8` UNITS |
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| 13802. |
The refractive index of flints glass prism for C, D and F lines are 1.790,1.795 and 1.805 respectively. Find the dispersive power of the flint glass prism. |
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Answer» SOLUTION :`mu_c=1.790, mu_D=1.795 and mu_F=1.805` `OMEGA=(mu_F-mu_c)/(mu_D-1)=(1.805-1.790)/(1.795-1)=0.015/0.795=0.1887` |
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| 13803. |
A point performs simple harmonic oscillation of period T and the equation of motion is given by x= a sin(omega t+(pi)/(6)). After the elapse of what fraction of the time period, the velocity of the point will be equal to half of its maximum velocity? |
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Answer» `(T)/(3)` `root(OMEGA)(a^(2)-x^(2))= (a omega)/(2)` `a^(2)- x^(2) = (a^2)/(4)` `therefore a^(2)- (a^2)/(4)= x^(2)`. `therefore x= (sqrt(3))/(2)a` Now, `x= a sin (omega t+(pi)/(6))` `(sqrt(3))/(2)a= a sin (omega t+(pi)/(6))` `therefore (sqrt(3))/(2) = sin (omega t+(pi)/(6))` `therefore omega t+(pi)/(6)= (pi)/(3)` `therefore omega t= (pi)/(3)-(pi)/(6)` `therefore omega t= (pi)/(6)` `therefore t= (pi)/(6)xx (T)/(2pi)` `therefore t= (T)/(12)`. |
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| 13804. |
A glass tube of diameter 0.06cmis dipped in methyl alcohol, if angle of contact is 0^(@), density of methyl alcohol 0.8gcm^(-3) and surface tension 0.023Nm^(-1) the height to which it rises in the tube is |
| Answer» ANSWER :B | |
| 13805. |
Screw gauge can be used to determine. |
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Answer» The THICKNESS of a thin glass plate |
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| 13806. |
A floating body, in a liquid, sinks to the bottom when heated. Why? |
| Answer» SOLUTION :Before HEATING the body is floating in a liquid since the density of the body is LESS than the density of the liquid. On heating if the decrease in density of the liquid is more than the decrease in density of the floating body, the body SINKS to the bottom. | |
| 13807. |
The dimensions of quantities in one or more pairs are same . Identify them |
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Answer» torque and force |
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| 13808. |
When I m long metallic rod is vertically dropped it strikes the solid metallic floor and 1.2 kHz. Find speed of this sound wave in the rod. |
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Answer» `600 m//s` `v = LAMDA F ` `= f _(min) lamda _(max)` `= f _(min ) xx 2L` `= 1.2 xx 10 ^(3) xx 2 xx 1` `= 2400 m//s` |
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| 13809. |
A flat car of mass m, starts moving to the right due to a constant horizontal force. Sand spills on the flat car from a stationary hopper. The velocity of loading is constant and equal to mu Kg/s. Find the time dependance of the velocity and the acceleration of the flat car in the process of loading. The friction is negligibly small. |
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Answer» Solution :`m(dv)/(dt)+ v(dm)/(dt) = F ` MASS of the car at any instant `= m_0 + mut ` `:. ( m_0 + mu t) (dv)/(dt) + v mu =F` `:. (dv)/(F- mu v) =(dt)/(m_0 + mu t)` Integrating , we have `log (F- mu v) = - log ( m_0 + mu t) + C` when t=0 , v=0 `:. log F = - log m_0 + C` `:. log (F-muv) = - log ( m_0 + mu t) + log F + log m_0` `:. (F_ mu v)/( F) = (m_0)/( m_0 + mu t)` `(mu v)/(F) = (mu t)/(m_0 + mu t)[" SINCE " if (a)/(b) = (C)/(d) ,(b-a)/(b) =(d-C)/(d)] or v=(FT)/(m_0 + mu t)` `:.` Acceleration `=(dv)/(dt) = (d)/(dt) ( Ft)/( m_0 + mu t) = ((m_0mut) F-F t mu)/((m_0 + mut)^(2))` `=(F m_0 + mu F t - mu Ft)/( (m_0 + mut)^(2)) = (F m_0)/(m_0^(2)(1 + (mut)/(m_0))^(2))=(F)/(m_0 (1 + (mut)/(m_0))` |
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| 13810. |
An observer can see through a pin hole the top end of a thin rid of height h, placed as shown in figure. The beaker height 3h and its radius h. When the beaker is filled with a liquid upto a height 2h, he can see the lower end of the rod. Find the reactive index of the liquid. |
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Answer» SOLUTION :`PQ=QR=2h` `THEREFORE anglei=45^@` `therefoe ST=RT=h=KM=MN` so `KS=sqrt(h^2+(2h)^2)=h sqrt5` `therefore sin r= h/(h sqrt5)=1/sqrt5` ` therefore MU= (sin i)/( sin r)= (sin 45^@)/(1//sqrt5)= sqrt(5/2)` 
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| 13811. |
A car is moving with a speed of 30 ms^(-1) on a circular path of radius 500 m . If its speed is increasing at the rate of 2 ms^(-2), the net acceleration of the car is |
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Answer» `3.6 MS^(-2)` |
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| 13812. |
Three objects each of mass .m. are placed at the corners of an equilateral triangle of side .l.. What is the gravitational force acting on each mass due to other masses is |
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Answer» `(Gm^2)/(SQRT(3l^2))` |
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| 13813. |
A particle is in clockwise uniform circular motion the direction of its acceleration is radially inward. If sense of rotation or particle is anti-clockwise then what is the direction of its acceleration? |
| Answer» SOLUTION :RADIAL INWARD | |
| 13814. |
At a height H from the surface of the earth , the total energy of a satellite is equal to the potential energy of a body of equal mass at a height of 3R from the surface of earth (R-Radius of the earth ) . The value of H is |
| Answer» Answer :B | |
| 13815. |
From the top of a tower a stone is thrown up which reaches the ground in time t_(1). A second stone thrown down with the same speed reaches the ground in a time t_(2). A third stone released from rest from the same location reaches the ground in a time t_(3). then: |
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Answer» `(1)/(t_(3)) = (1)/(t_(1)) + (1)/(t_(2))` `H = u t_(2) + (1)/(2) g t_(2)^(2)` `H = (1)/(2) g t_(3)^(2)` |
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| 13816. |
Five persons A, B, C, D & E are pulling a cart of mass 100kg on a smooth surface and cart is moving with acceleration 1 m//s^2in east direction. When person 'A' stops pulling, it moves with acceleration 1m/s in the west direction. When person 'B' stops pulling, it moves with acceleration 24m//s^2in the north direction. The magnitude of acceleration of the cart when only A & B pull the cart keeping their directions same as the old direction is |
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Answer» `26 m//s^2` |
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| 13817. |
Two identical mass M moving with velocity u_1 and u_2 collide perfectly inelastically. The loss in kinetic energy is |
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Answer» `M/2 (u_2 - u_1)^2` `DELTA K = 1/2 (m_1m_2)/((m_1 + m_2)) (u_1 - u_2)^2` Here, `m_1 = m_2 = M` `:. Delta K = 1/2(M XX M)/((M + M)) (u_1 - u_2)^2 = M/4 (u_1 - u_2)^2` . |
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| 13818. |
Ten coins are placed on top of each other on a horizontal table. If the mass of each coin is 10g and acceleration due to gravity is 10 ms^(-2), what is the magnitude and direction of the force on the 7^(th) coin (counting from the bottom) due to all the coins above it ? |
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Answer» 0.3 N downwards |
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| 13819. |
A circular disc is rotating about its own axis at an angular velocity omega. If P is exact midpoint of dise between axis and rim of disc then angular velocity of Pis |
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Answer» `(omega)/(2)` |
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| 13820. |
A pebble is thrown vertically upwards from a bridge with an height of the bridge is : |
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Answer» 19.6 m HENCE, they will reach to GROUND at same TIME. |
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| 13821. |
A soldier fires bullets, each of mass 50 g, from his automatic rifle with a velocity of 1000 m/s. If he can bear a maximum force of 200 N on his shoulder, find the number of bullets which he can fire in a second. |
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Answer» SOLUTION :Themomentumimparted ton bullets persecond = (nmv-0 )= nmv(momentumofa bullet=MV ) `=nmv = (DP)/(dt )` `=:. , 200nmv ` `n = (200)/( mv)= (200)/( 50 XX 10^(-3)xx 10^(3))` `:., n= 4` |
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| 13822. |
A particle of mass M is placed at the centre of a uniform spherical shell of mass 2M and radius R. The gravitational potential on the surface of the shell is: |
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Answer» `-(GM)/R` |
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| 13823. |
A force of (2 hat(i) + hat(j) + 2hat(k))N is acting on a body, shifted it from (1, 2,0) to (2, 3,4) in the same frame of reference Angle between the displacement and force |
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Answer» `"COS"^(-1) (11)/(9 sqrt2)` |
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| 13824. |
A force of (2 hat(i) + hat(j) + 2hat(k))N is acting on a body, shifted it from (1, 2,0) to (2, 3,4) in the same frame of reference The vector component of the force along the displacement of the body |
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Answer» `(5)/(sqrt6) (2 hat(i) + hat(j) + 3HAT(k))` |
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| 13825. |
A body starts falling from the top of a smooth sphere of radius r. What angle does the bodysubtend at the centre of the sphere when it just loses contact with the sphere and what willbe its velocity then ? |
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Answer» Solution :Let the ANGLE subtended by the body atthe centre of the sphere when it loses contact with the sphere be `theta`. The body just loses contact with the sphere when the component of gravitational force cannot provide the NECESSARY centripetal force to the body so that it can continuein the circular path. So, according to Fig .1.36.  `mg cos theta =(mv^2)/r...(1)` (V= velocity of the body WHENIT just loses contact with the sphere) According tothe principle of conservation of energy, `1/2 mv^2 =mgr (1- cos theta)` [`OC=OA=r, OB=r cos theta BC=OC-OB=r-r cos theta=r(1-cos theta)]` or, `v^2 =2gr(1-cos theta)` From equations(1) and (2) we get, `mg cos theta =m/r*2gr(1-cos theta)` or, `cos theta =2(1-cos theta)` `or, cos theta =2/3 or, theta=cos^(-1)"" 2/3=48.2 ^@` again ,`v^2=2gr(1-cos theta)=2gr(1-2/3)=2/3gr` `therefore v=sqrt(2/3)gr`. |
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| 13826. |
For a projectile .R. is range and .H. is maximum height |
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Answer» a-g, b-h, C-e, d-f |
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| 13827. |
The displacement time graphs of two moving particles make angles of 30^(@) and 45^(@) with the x-axis. The ratio of the two velocities is |
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Answer» `SQRT(3):1` |
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| 13828. |
In the adjacent figure there is a cube having a smooth groove at an inclination of 30° with horizontal in its vertical face. A cylinder A of mass 2kg can slide freely inside the grove. The cube is moving with constant horizontal acceleration A_(0) parallel to the shown face, so that the slider does not have acceleration along horizontal. |
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Answer» The normal reaction acting on cube is ZERO |
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| 13829. |
In an experiment of simple pendulum a student made several observations for the periodof oscillation. His readings turned out to be 2.63s, 2.56s, 2.42s, 2.71s, and 2.80s. Find (i) mean time period of oscillations or most accurate value of time period, (ii) absolute error in each reading (iii) mean absolute error (iv) fractional error and (v) percentage error. |
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Answer» Solution :(i) The MEAN time period of oscillation T=`(2.63 +2.56+2.42+2.71+2.80)/(5)` `=(13.12)/(5)s = 2.624 s~~ 2.62s` (ROUNDED off to 2nd decimal PALCE ) (ii) Taking 2.62s as the true value the absolute errors (true value - measured value ) in the five READINGS are, (2.62 - 2.63) s= -0.01s, (2.62-2.56) s = 0.06s, (2.62 - 2.42)s = 0.20s, (2.62 - 2.71)s = -0.09 s and (2.62-2.80) s =-0.18s (iii) The (maximum) mean absolute error is `(deltaT)_("max")=(0.01+0.06+0.20+0.09+0.18)/(5)`s `=(0.54)/(5)s = 0.108 s ~~ 0.11 `s (iv) The (maximum) fractional error is `((deltaT)/(T))_("max")=(0.11s)/(2.62s)~~0.04` (v) The maximum percentage error is `((deltaT)/(T))_("max")xx100 = 0.04 xx100 =4%` `:.` The value of T should be written as (`2.62 pm 0.11)`s |
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| 13830. |
The accelerated train is an example for |
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Answer» INERTIAL FRAME |
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| 13831. |
A fan produces a feeling of comfort because |
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Answer» A fan SUPPLIES COOL air |
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| 13832. |
A stone is dropped from a height h. Another stone is thrown simultaneously in the vertical direction so as to rise to a height 4h. How much later would the two stones cross eachother? |
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Answer» `SQRT((H)/(8G))` |
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| 13833. |
A uniform sheet of steel is cut into the shape as shown. Compute the X and Y co-ordinates of the centre of mass of the piece. (Width of each plate is 10 cm) |
Answer» SOLUTION : `(X_(CM),Y_(cm))=(22.5,17.5)` |
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| 13834. |
A horizontal converyor belt moves with a constant velocity V. A small block is projected with a velocity of 6 m/s on it in a direction opposite to the direction of motion of the belt. The block comes to rest relative to the belt in a time 4s. mu = 0.3, g = 10 m//s^(2). Find V |
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Answer» SOLUTION :`|vec(V)_(b,c)|=V_(b)+V_(c )=6+V` `f=mu mg = 0.3 xx m xx 10 = 3m` Retardation a `= (3m)/(m)=3m//s^(2)` `u_(R )=6+V, V_(r )= 0, t = 4` SEC `a_(r ) = - 3 ms^(-2)` `V_(r )= u_(r )+a_(r )t` `0=(6+V)-3xx4` V = 6 m/s. |
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| 13835. |
Centre of mass of the body practically coincides with: |
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Answer» its CENTRE of gravity |
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| 13836. |
Of the following substances, which one can be lifted upwards with more ease with the help of a hydrogen-filled balloon? |
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Answer» 1 KG steel |
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| 13837. |
Find (1.03)^(1//3) |
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Answer» SOLUTION :`(1+0.03)^(1//3) (x=0.03 &N=(1)/(3))` `~=1+(0.03)/(3)=1.01` |
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| 13838. |
Define and explain displacement |
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Answer» Solution :The change in position of particle in particular interval of time is called displacement. Let `x_(1) and x_(2)`be the positions of an object at time `t_(1) and t_(2)` Its displacement, denoted by `Deltax` in time `Delta t (t_(2) -t_(1))`is GIVEN by the difference between the FINAL and initial positions. `Delta X =x_(2) -x_(1)` We use the Greek letter delta (A) to denote change in a quantity If `x _(2) gt x _(1)Delta x`is positive, and if `x_(2) lt x _(1) Delta x` is negative. Displacement has both magnitude and direction In one-dimensional motion there are only two DIRECTIONS (backward and forward, upward and downward) in which an object can move. These two directions can easily be SPECIFIED by + and-signs. For example, displacement of the car in moving from O to P is, `Delta x x_(2) - x_(1)` (+ 360 m) - Om = + 360 m The displacement has a magnitude of 360 m and is directed in the positive x-direction as indicated by the + sign. Similarly, the displacement of the car from P to is 240 m - 360 m - 120 m. The negative sign indicates the direction of displacement.. 
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| 13839. |
A particle, (a mud pellet, say) of mass m strikes a smooth stationary wedge of mass M with a velocity Vo, at an angle theta with horizontal. If the collision is perfectly inelastic, find the : a) velocity of the wedge just after the collision. b) change in KE of the system (M + m) in the collision. |
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Answer» Solution :(a) Let the system (M +m) MOVE as a single mass with a VELOCITY v. Conserving the momentum of the system HORIZONTALLY , we have `mv _(0) cos theta = (M+m) v` This gives `v = (mv _(0) cos theta)/( M +m)` (b) The change in KE of the system is `Delta K = 1/2 (M +m) v ^(2) - 1/2mv _(0) ^(2) ` where `v = (mv _(0) cos theta)/( M +m)` This gives `Delta K = ((M+m))/( 2) (( mv _(0) cos theta ) /(M + m )) ^(2) -1/2 mv _(0) ^(2)` `=- ( mv _(0) ^(2))/( 2) (1- (m)/( M + m ) cos ^(2) theta ) = - ((M + m SIN ^(2) theta) mv _(0) ^(2))/( 2 (M + m ))` 
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| 13840. |
Calculate the quantity of heat required to convert 10^(2) kg of ice at 0^(0)C to steam at 100^(0)C. (Given L_("ice") =0.335 xx 10^(6) JKg^(-1), L_("steam") = 2.26 xx 10^(6) JKg^(-1). Specific heat of water ="4186J Kg"^(-1) K^(-1). |
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Answer» Solution :QUANTITY of heat required to CONVERT ice into WATER at `O^0C=, (Q_1)= mL, =10^(-2) xx 0.355 x 10^(-6)=3350J` Quantity of heat required to raise the temperature of water from `0^@C" to "100^@C`. `(Q_2) =mS(12-1) = 10^(-2)xx 4186 xx100= 4186J` Quantity of heat required to convert water into steam at `100^@C` `(Q_3)= mL_2, =10^(-2) xx 2.26 xx 100= 22600J` Total amount of heat required `(Q) =Q_1+Q_2+Q_3= 3350 + 4186 +22600= 30136J` |
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| 13841. |
If a particle is executing SHM on a straight line. A and B are two points at which its velocity is zero. It passes through a certain point P (AP lt PB) at successive intervals of 0.5 sec and 1.5 sec with a speed of 3m/s. The maximum speed of the particle is |
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Answer» 3m/s |
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| 13842. |
Consider a cylinder of mass M and radius R luing on a rough horizontal plane. It has a plank lying on it top as shown in figure. A force F is applied on the plank such that plank moves causes the cylinder to roll. The plank always remains horizontal. There is no slipping at any point of contact then |
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Answer» the ACCELERATION of the cylinder is `(4Fcostheta)/(8m+3M)` |
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| 13843. |
A massive ball moving with speed v collides head on with a fine ball having mass very much smaller than the mass of the first ball. The collision is elastic. Then immediately after the impact, the second ball will move with a speed approximately equal to |
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Answer» v |
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| 13844. |
A rider on a horse back falls forward if the horse suddenly stops. This is due to |
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Answer» the large WEIGHT of the horse |
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| 13845. |
What is called natural convection ? |
| Answer» Solution :The CONVECTION in which, heat moves with motion of material DUE to density DIFFERENCE is called NATURAL convection. | |
| 13846. |
A piece of cork whose weight is 19 gm is attached to a bar of silver weighing 63 gm. The two together just not in water. The specific gravity of silver is 10-5. Find the specific gravity of the cork. Density of water = 1 gm cm ^(-3) |
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Answer» SOLUTION :Density of SILVER ` = 10.5 gm cm ^(-3)` `"" ` MASS of cork = `m_1= 19 gm ` `""`Mass of silver `= m_2 = 63 gm` Total mass of FLOATING body = ` m_1+ m_2 = 82 gm ` Mass of floating body= mass of water DISPLACED ` therefore ` Mass of displaced = ` ("mass " )/( " density")= (82)/(1) = 82 cm ^(3)` Volume of silver `= (" mass")/(" density")= ( 63)/( 10.5) = 6 cm ^(3)` Volume of cork = Total Volume - Volume of silver ` "" = 82- 6= 7^(2)cm ^(3)` Density of cork = ` (" mass ")/( " volume")= (19)/(76) = 0.25 gm cm ^(-3)` Specific gravity = ` (" density of cork")/( " density of water ")= (0.25 gm cm ^(-3))/( 1 gm cm ^(-3))= 0.25` |
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| 13847. |
In the arrangement shown in the figure, A is a jar half filled with water and half filled with air. It is fitted with a leak proof cork. A tube connects it to a water vessel B. Another narrow tube fitted toA connects it to a narrow tube C via a water monometer M. The tip of the tube C is just touching the surface of a liquid L. Valve V is opened at time t = 0 and water from vessel B pours down slowly and uniformly into the jar A. An air bubble develops at the tip of tube C. The cross sectional radius of tube C is r and density of water is rho. The difference in height of water (h) in the two arms of the manometer varies with time 't' as shown in the graph. Find the surface tension of the liquid L. |
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Answer» |
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| 13848. |
By how much amount should the temperature of a steel rod be increased so that its length increases by 0.5%? The coefficient of linear expansion of steel is 1.2 xx 10^(-5) C^(-1) |
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Answer» Solution :`(DELTA l)/(l) = (0.5)/(100) , ALPHA = 1.2 XX 10^(-5) ""^@C^(-1)` `Delta T = ?` ` Delta l = alpha Delta T` ` Delta T = (Delta l)/(l alpha) = (0.5)/(100 xx 1.2 xx 10^(-5) ) = 416.7^@C` |
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| 13849. |
A simple apparatus for demonstrating resonance in an air column is depicted in Fig. 7.59.A vertical pipe open at both ends is partially submerge in water , and a tuning fork vibration at an unknown frequency is placed near the top of the pipe . The length L of the air column can be adjusted by moving the pipe vertically . The sound waves generated by the fork are reinforced of the pipe . For a certain pipe , the smallest value of L for which a peak occurs in the intensity is 9.00 cm. a. What is the frequency of the tuning fork ? b. What are the values of L for the next two resonance conditions? |
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Answer» Solution :Consider how this problem differs from the PRECEDING problem. In the culvert , the length was fixed and the air column was presented with a mixture of many frequencies . The pipe in this exampleis presented with one single frequency from the tuning fork , and the lenght of the pipe is varied until RESONANCE is achieved. This EXAMPLE is a simple substitution problem . Although the pipe is open at its lower end to allow the water to enter , the water's surface acts like a barrier . Therefore , this setup can be modelled as an air column closed at one end . The fundamental frequency for `L =0.090 m`: `f_(1) = (V)/( 4L) = ( 343 m//s )/(( 4( 0.090 m)) = 953 Hz` Because the tuning fork causes the air column to resonate at this frequency , this frequency must also be that of the tuning fork. b. Use equation `v = lambda f` to FIND the wavelength of the sound wave from the tuning fork. `lambda = (v)/(f) = ( 343 m//s)/( 953 Hz) = 0.360 m` Notice from Fig. 7.59 (b) that the length of the air column for the second resonance is ` 3 lambda//4`. `L = 3 lambda//4 = 0.270 m` Notice from Fig. 7.59(b) that the length of the air column for the third resonance is ` 5 lambda//4` `L = 5 lambda//4 = 0.450 m` |
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| 13850. |
A ball is dropped vertically, from a height h above the ground. It hits the ground and bounces up vertically to a height h/2. Neglecting subsequent motion and air resistance, its velocity v varies with the height h as |
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