InterviewSolution
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InterviewSolution
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.
| 1. |
A particle moves along a curve of unknown shape but magnitude of force `F` is constant and always acts along tangent to the curve.ThenA. `F` may be conservativeB. `F` must be conservativeC. `F` may be non conservativeD. `F` must be non-conservative |
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Answer» Correct Answer - D Since,work done by this force depends upon path followed by the particle betweeb two. Hence, the force must be non.coservatice. |
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| 2. |
Which of the following velocity-time graphs shows a realistic situation for a body in motion ?A. B. C. D. |
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Answer» Correct Answer - B Except graph (b), other graphs shows the motion of the body with more than one velocity which is not possible for realistic situation. |
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| 3. |
As shown in figure in a simple barmonic motion obscillator having identical four springs has time period A. `T = 2 pi sqrt((m)/(4 k))`B. `T = 2 pi sqrt((m)/(2 k))`C. `T = 2 pi sqrt((m)/( k))`D. `T = 2 pi sqrt((m)/(8 k))` |
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Answer» Correct Answer - c `k_(1)` and `k_(2)` parallel and `k_(3)` and `k_(4)` are parrallel .The two combnations an in same with each other `k_(1) + k_(2) = 2k` `k_(3) + k_(4) = 4k` `(1)/(k_(eq)) = (1)/(2k) + (1)/(2k)` `k_(eq) = k` `T = 2 pi sqrt((m)/(k))` |
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| 4. |
Assertion: The error in the measurement of radius of sphere is `0.3%`. The permissible error in its surface area is `0.6 %`. Reason: The permissible error is calculated by the formula `(Delta A)/(A) = (4 Delta r)/( r)`.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - C `A = 4 pi r^2` (error will not be involved in constant `4 pi`) Fractional error `(Delta A)/(A) = (2 Delta r)/(r )` `(Delta A)/(A) xx 100 = 2 xx 0.3 % = 0.6 %` but `(Delta A)/(A) = (4 Delta r)/( r)` is false. |
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| 5. |
The dimension of torqueA. `[ML^(2)T^(-3)]`B. `[ML^(-1)T^(1)]`C. `[ML^(2)T^(-2)]`D. `[MT^(-2)]` |
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Answer» Correct Answer - c The moment of a force on the torque about at axis reration is equal to the product of the magnitude of the force (F) and the perpendicular ® of the line of action of force from the axis of ration `:. tau = F xx t` Dimensions of energy `=` Dimensions of force`=` Dimensions of distance `= [MLT^(-2)] xx [L] = [ML^(2)T^(2)]` |
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| 6. |
Assertion: The change in air pressure effects the speed of sound. Reason: The speed of sound in gases is alphaortional to the square of pressure.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - D The speed of sound in gaseous medium is gien by `v = sqrt((gamma p)/(rho))` …(i) At constant temperature `PV =` constant …(ii) If `V` is the volume of one mole of a gas, then density of gas `rho = (M)/(V)` or `V = (M)/(rho)` Where `M` is the molecular weight of the gas. `:.` Eq.(ii) becomes `(pM)/(rho)` = constant or `(p)/(rho)` = constant as `M` is a constant Therefore, from Eq. (i), we have `v = "constant" xx sqrt(gamma)` Thus, change inair pressure does not effect the speed of sound. Reason is clear from Eq. (i). |
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| 7. |
A gun fires a bullet of mass `50 g` with a velociy of `30 m//s`. Due to this, the gun is pushed back with a velocity of `1 m//s`, then the mass of the gun is :A. `1.5 kg`B. `5.5 kg`C. `0.5 kg`D. `3.5 kg` |
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Answer» Correct Answer - A Momentum remains conserved in the process. If no external force acts upon a system of two(or more) bodies then the toatal momentum of the system remains constant. Hence, momentum before collision= momentum after collision `m_(1)u_(1)=m_(2)v_(2)` Given, `u_(1)=1 m//s,m_(2)=0.05 kg,v_(2)=0.05 kg,v_(2)= 30m//s` `implies m_(1)xx1= 0.05xx30` `implies m_(1)= 1.5 kg` |
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| 8. |
Dimension of resistance in an elecatrical circuit, in terms of dimension of mass `M`, of length `L`, of time `T`, and of current `I`, would beA. `[ML^(2)T^(3)I^(-I)]`B. `[ML^(2)T^(2)]`C. `[ML^(2)L^(-1)]`D. `[ML^(2)T^(3)]` |
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Answer» Correct Answer - D Resistance `R=("potential di fference")/("Current")=V/i=W/(qi)` (`:.` Potential difference is equal to work done per unit charge) So, Dimension of `R` `=(["Dimension of work"])/(["Dimension of charge"]["Dimension of current"])` `=([ML^(2)T^(-2)])/([IT][I])=[ML^(2)T^(-3)I^(-2)]` |
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| 9. |
Assertion: In taking into account the fact that any object which floats must have an average density less than that of water, during World war `I`, a number of cargo vessels are made of concrete. Reason : Conctere cargo Vessel were filled with air.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is false.D. If both aseertion and reason are false. |
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Answer» Correct Answer - A The density of concreate of course, is more than that of water and a block of concrete will sink like a stone,if dropped into water. Concreate cargo were filled with air and as such, average density of cargo Vessels `=("mass o f concreate + mass of air")/("volume of concreate + volume of air")` It follows that the avarage density of cargo vessels must be less than that of water. As a result the concreate cargo vessels did not sink. |
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| 10. |
A spherical drop of water has `1mm` radius. If the surface tension of water is `70xx10^(-3)`N//m, then the difference of pressure between inside and outside of the spherical drop is:A. `140 N//m^(2)`B. `14 N//m^(2)`C. `35 N//M^(2)`D. None of the above |
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Answer» Correct Answer - A Let the radius of drop be `R` and surface tension water be `T`. Then the excess pressure of difference pressure between inside and outside of the spherical drop is given by `P=(2T)/(R )` Given, `R=1 mmm= 1xx10^(-3)m` `T= 70xx10^(-3)N//m` `:. P=(2xx70xx10^(-3))/(1xx10^(-3))= 140 N//m` |
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| 11. |
The volume of a gas is reduced adibatically to `(1//4)` of its volume at `27^(@)C` if `y = 1.4` The new temperature will beA. `300 xx (4)^(0.4) K`B. `150 xx (4)^(0.4) K`C. `250 xx (4)^(0.4) K`D. None of these |
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Answer» Correct Answer - a For adiabatic change the relation between temperature and volume is `TV^(gamma-1) = "constant"` where `gamma` is ratio of specific heats of the gas Given `T _(1) = 27 + 273 = 300 K , V_(1)= V, V_(2) = (V)/(4)` `T_(1)V_(1)^(gamma- 1) = T_(2) V_(2)^(gamma - 1)` `rArr T_(2) = ((V_(1))/(V_(2)))^( gamma - 1) xx T_(1)` `T_(2)= ((V)/(V//4))^(1.4 - 1) xx 300` `T_(2) = (4)^(0.4) xx 300 xx (4)^(0.4) K`. |
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| 12. |
Assertion : The root mean sguar and most probable speed of the molecules in a gas are the same Reason : The Maxwell distribation for the speed of molecules in a gas is symentricalA. If both assertion and reason are true and the reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true statement but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - d Root mean squre speed of moleclues in a gas is define as the equal root of mean of space of the speed of difference i.e. `v_(min) = sqrt(((v_(1)^(2) + v_(2)^(2) + ......)/(N))` `or v_(min) = sqrt((3RT)/(M))` (according to kinetic theory gases) while most position speed is the speed which maximum number of molecles in a gas have at common temperature and is given by `v_(m) = sqrt((2RT)/(M))` It is abvious that `v_("rms") gt v_("mp").` Also maxwell damenation for the speed of moleculas in a gas is any numerical hence option (d) is true. |
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| 13. |
Assertion: In adiabatic compression, the internal energy and temperature of the system get decreased. Reason: The adiabatic compression is a slow process.A. If both assertion and reason are true and the reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true statement but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - c In adiabastic process, there is no exchange of heat between the system and the surrounding .This can be position if the gas under adibatic process is allowed to expand an compressed very quickly Thus, it is a quik process when the gas is comprecessed adatically , the heat preshers canot ascepe is the result surrending through the including walls , the temperature of the gas and hence the initial energy increases hence option (c) is correct. |
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| 14. |
The velocity of a particle moving in the `x-y` plane is given by `(dx)/(dt) = 8 pi sin 2 pi t and (dy)/(dt) = 5 pi sin 2 pi t` where, `t = 0, x = 8 and y = 0`, the path of the particle is.A. a straight lineB. an ellipseC. a circleD. a parabola |
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Answer» Correct Answer - B `y-x` graph gives the shape of path of particle `(dx)/(dt) = 8 pi sin 2 pi t` `int_8^x dx = int_0^t 8 pi sin 2 pi t dt` `x - 8 = -(8 pi)/(2 pi)[cos 2 pi]_0^t` `x - 8 = 4(1 - cos 2 pi t) rArr cos 2 pi t = (x - 12)/(4)` `(dy)/(dt) = - 5 pi cos 2 pi t` `int_0^y dy = 5 pi t int_0^t cos 2 pi t rArr y = (5)/(2) sin 2 pi t` `((x - 12)/(4))^2 + (y^2)/((5/2)^(2)) = 1 rArr ((x - 12)^2)/((4)^2) + (y^2)/(5/((2)^(2))) = 1`. |
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| 15. |
A string in a musical instrument is `50 cm` long and its fundanmental frequency is `800 Hz` If the frequency of `1000 Hz` is to be produced then required length of spring isA. `37.5 cm`B. `40 cm`C. `50 cm`D. `62.5 cm` |
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Answer» Correct Answer - b The frequency producted in a string of length of L mass per unit length m , and tension T is `n = (1)/(2l) sqrt((T)/(m))` Given `l_(1) = 50 cm , n_(1) = 800 Hz` M and `n_(2) = 1000 Hz` `:. (n_(1))/(n_(2)) = (l_(2))/(l_(1))` `rArr l_(2) = (n_(1)l_(1))/(n_(2)) = (800 xx 50)/(100) = 40 cm` |
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| 16. |
A neutron makes is a head - on elastic collision with a stationary deuteron The fraction energy loss of the neutron in the collision isA. `16//81`B. `8//9`C. `8//27`D. `2//3` |
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Answer» Correct Answer - b Apply law of coservation of momentumm and coservation of energy Let the two balls of mass `m_(1)` and `m_(2)` collide each other electrically with volicities `u_(1)` and `u_(2)`. Their velocities become `v_(1)` and `v _(2)` after the collision. Applying conservation of linear momentum, we get `m_(1)u_(1)+m_(2)u_(2)= m_(1)v_(1) +m_(2)v_(2)` ...(1) Also from conservation of kinetic energy `(1)/(2)m_(1)u_(1)^(2) + (1)/(2)m_(2)u_(2)^(2) = (1)/(2) m_(1)v_(1)^(2) + (1)/(2) m_(2)v_(2)^(2)`...(2) Solving Eqs `(1)` and `(2)` we get `v_(1) = ((m_(1) - m_(2))/(m_(1) + m_(2))) u_(1) + ((2m_(2))/(m_(1) + m_(2))) u_(2)`...(3) `and v_(2) = ((m_(2) - m_(1))/(m_(1) + m_(2))) u_(2) + ((2m_(2))/(m_(1) +(m_(2))) u_(1))`...(4) On taking apporaximate value the mass deterons is twice the mass of neutron Given `u_(1) = u , u_(2) = 0 , m_(1)=m,m_(2)= 2m` Velocity of neutron, `v_(1) = ((m - 2m)/(m + 2m)) u = -(u)/(3)` Velocity of deuteron, `v_(2) = ("2mu")/(m + 2m) = (2)/(3) u` Fractional energy loss `= ((1)/(2)m u^(2) - (1)/(2)m( -(u)/(3))^(2))/((1)/(2)m u^(2))` `= 1 - (1)/(9) = (8)/(9)` |
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| 17. |
The magnetic moment has dimensions ofA. `[LA]`B. `[L^(2)A]`C. `[LT^(-1)A]`D. `[L^(2)T^(-1)A]` |
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Answer» Correct Answer - b Magnetic moment of a current carrying coul is definet as the product of current in the cell with the are of unit in vector from That is `overset(vec)(M) = 1 overset(vec)(r) A` Thus, dimensions of `[M] = [A] [L^(2)]` `= [L^(2)A]` |
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| 18. |
As what temperature the speed of sound in air will because double of its value at `27^(@)`?A. `54^(@)C`B. `627^(@)C`C. `927^(@)C`D. `327^(@)C` |
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Answer» Correct Answer - c Speed of sound is given by `v = sqrt((gamma P)/(A))` ..(1) where rho is ratio of specific heat P is pressure and d is density Also `PV = RT,` where `V` is volume R gas constant and T is temperature `P (m)/(d) = RT` `rArr (P)/(d) = (RT)/(m)` putting the value in Eqs (i) we get `v = sqrt((gamma RT)/(m))` Given ` v_(1) = v, T_(1) = 27 + 273= 330 K` `v_(2) = 2v` `:. (v_(1))/(v_(2)) = (v)/(2v) = sqrt((T_(1))/(T_(2))) = sqrt((300)/(T_(2)))` `rArr T_(2) = 4 xx 300 = 1200 K` `rArr T_(2) = (1200 - 273)^(@)C = 927^(@)C ` |
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| 19. |
The tension in a piano wire is `10 N`. The tension ina piano wire to produce a node of double frequency isA. `20 N`B. `40 N`C. `10 N`D. ` 120 N` |
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Answer» Correct Answer - B For mass `m` per unit length of wire and tension `T`, the frequency of note emitted by the wire is `n=1/(2l)sqrt(T/m)` where `l` length of wire. Given `T_(1) = 10N, n_(1)=n` and `n_(2)=2n` `(n_(1))/(n_(2))=sqrt((T_(1))/(T_(2)))` `implies T_(2)= 10xx4= 40N` |
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| 20. |
For inclestic collsion between two spherical rigid bodiesA. the total kinetic energy is conservedB. the total machanical energy is not conservedC. the linearmomentum is not conservedD. the linearmomentum is conserved |
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Answer» Correct Answer - d In an ineclastic collision , the particles do not regain their shape and size completely after collision some fraction of machanical energy is retained by the colliding particle in the form of deformation piotential energy This the kinetic energy of particles no loeger remain conservad , Howevers in the absonce of external force law of conservation of linear momentum still holds good |
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| 21. |
A string with a frequency `n` and `B` string with a frequency `1//8` that of `A`. If the energy remains the same and the amplitude of `A` is `a`, then amplitude of `B` will beA. `2a`B. `8a`C. `4a`D. `a` |
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Answer» Correct Answer - B The energy of a wave travelling with velocity `v` is given by `E= 1/2 mv^(2)` Also `v= amplit udexxangu lar ve locity` `implies v= a omega` `:. E= 1/2ma^(2)omega^(2)` Also `omega= 2pin` Where `n` is frequency. `:. E= 1/2 ma^(2)(2pin)^(2)` `implies E prop a^(2)n^(2)` It is given that energy remains the same. Hence, `E_(A)=E_(B)` `:. (a_(A)/(a_(B)))^(2)=(n_(B)/(n_(A)))^(2)` Given,`n_(A)=n,n_(B)=n/8` `:. (a_(A))/(a_(B))=(n//8)/(n)=1/8` `implies a_(B)=8a_(A)= 8a` |
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| 22. |
if the vectors `P = alphahati+alphaj+3hatk` and `Q = alphahati - 2 hatj - hatk` are perpendicular to each other, then the positive value of `alpha` isA. zeroB. `1`C. `2`D. `3` |
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Answer» Correct Answer - d The scalar praduct of two vectors is given by `overset(vec)(P) .overset(vec)(Q) = PQ cos theta` where `theta` is angle between them Given , `hatP = a hati + a hat j + 3hatk,vecQ = a hati - 2 hatj - hatk` `theta 90^(@)` `:. vecP . vecQ = PQ cos 90^(@) = 0` `(a hati - hatj -3 hatk).(a hati - 2 hatj - hatk) = 0` `a^(2) - 2a - 3 = 0` `rArr a = (2 +- sqrt(4 + 12))/(2) = 3 = 1` Hence position value of a is `3`. |
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| 23. |
Two ballons are filled one with pure he gas and the other by six repectively If the pressure and temperature of these bellows are same then the number of molecales per unit volum isA. mass in the He filled balloopB. mass in both filled balloopC. mass in air filled balloopD. in the ratio of `1.4` |
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Answer» Correct Answer - b Ideal gas equation can be written as `PV = pi RT`…(i) In this equation `n =` number of moles of the gas `P =` pressure of the gas `V =` volume of the gas `R =` universal gas constant and `T =` Temperature of the gas From Eqs (i), we have `(n)/(V) = (P)/(RT) = "constant"` So at constant pressure and temperature all gases will contain equal number of melecules per unit volume. |
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| 24. |
If the terminal speed of a sphere of gold (density `=19.5kg//m^3`) is `0.2m//s` in a viscous liquid (density `=1.5kg//m^3`), find the terminal speed of a sphere of silver (density `=10.5kg//m^3`) of the same size in the same liquidA. `0.4 m//s`B. `0.133 m//s`C. `0.1 m//s`D. `0.2 m//s` |
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Answer» Correct Answer - C Terminal speed of spherical body in a viscous liquid is given by `v_T = (2 r^2(rho - sigma)g)/(9 eta)` where `rho` = density of substance of body and `sigma` = density of liquid From given data `(v_T(Ag))/(v_T(Gold)) = (rho_(Ag) - sigma_l)/(rho_(Gold) - sigma_l)` `rArr v_t (Ag) = (10.5 - 1.5)/(19.5 - 1.5) xx 0.2 = (9)/(18) xx 0.2` =` 0.1 m//s`. |
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| 25. |
Two bodies of mass `m_(1) and m_(2)` are initially at rest placed infinite distance apart. They are then allowed to move towards each other under mutual gravitational attaction. Show that their relative velocity of approach at separation r betweeen them is `v=sqrt(2G(m_(1)+m_(2)))/(r)`A. `[2G((m_1 - m_2))/(r)]^(1//2)`B. `[(2G)/(r)(m_1 - m_2)]^(1//2)`C. `[(r)/(2G(m_1 m_2))]^(1//2)`D. `[(2G)/(r) m_1 m_2]^(1//2)` |
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Answer» Correct Answer - B Apply the principle of conservation of momentum and conservation of energy. Let velocities of these masses at `r` distance from each other be `v_1` and `v_2` respectively. By conservation of momentum `m_1 v_1 - m_2 v_2 = 0` `rArr m_1 v_1 = m_2 v_2` ...(i) By conservation of energy Change in `PE` = change in `KE` `(G m_1 m_2)/( r) = (1)/(2) m_1 v_1^2 + (1)/(2) m_2 v_2^2` `rArr (m_1^2 v_1^2)/(m_1) + (m_2^2 v_2^2)/(m_2) = (2 Gm_1 m_2)/( r)` ...(ii) On solving Eqs. (i) and (ii) `v_1 = sqrt((2 Gm_2^2)/(r(m_1 + m_2))) and v_2 = sqrt((2 Gm_1^2)/(r(m_1 + m_2)))` `:. v_(app) = |v_1|+|v_2|= sqrt((2 G)/(r)(m_1 + m_2))`. |
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| 26. |
The velocity of a bullet is reduce from `200 m//s` to `100m//s` while travelling through a wodden block of thickness `10 cm` Assuming it to be uniform, the retardation will beA. `15xx10^(4)m//s^(2)`B. `10xx10^(4)m//s^(2)`C. `12xx10^(4)m//s^(2)`D. `14.5m//s^(2)` |
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Answer» Correct Answer - A From equation of motion `V^(2)+u^(2)+2as`…(1) Where `v` is final velociy, `u` is initial velocity `a` is retardation and `s` is distance travel. Given, `v= 100 m//s, u= 200 m//s` `s=10 cm= 10xx10^(-2)` Putting the numerical values in Eq.(1), we have `:. (100)^(2)= (200)^(2)+2xx1xx10xx10^(-2)` `implies 30000= 2axx10^(-2)` `implies a=(30000)/(20xx10^(-2))` `= -15xx10^(4)m//s^(2)` Minus sign shows negatice acceleration i.e., retardation. |
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