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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. |
Assertion: In the relation`f = (1)/(2l) sqrt((T)/(m))`, where symbols have standard meaning , m represent linear mass density. Reason: The frequency has the dimensions linear of time.A. If both assertion and reason are true and reason is the correct explation of assertion.B. If both assertion 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 - b From `f = (1)/(2l)sqrt((T)/(m)) , f^(2) = (T)/(4l^(2)m)` or `m = (T)/(4l^(2)f^(2)) = ([MLT^(-2)])/(L^(2)T^(-2))= (M)/(L) = ("Mass")/("Length") =` linear mass density |
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| 2. |
Unit of `(CV)/(rho epsilon_(0))` are of (`C =` capacitance, `V =` potential, `rho =` specfic resistence and `epsilon_(0) =` permittivity of free space)`A. ChargeB. currentC. timeD. frequency |
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Answer» Correct Answer - 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"`. |
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| 3. |
A force `F` is given by `F = at + bt^(2)` , where `t` is time . What are the dimensions of `a and b`?A. `MLT^(-3) and ML^(2)T^(-4)`B. `MLT^(-3) and MLT^(-4)`C. `MLT^(-1) and MLT^(0)`D. `MLT^(-4) and MLT^(4)` |
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Answer» Correct Answer - b From the principal of dimensional bomogeneity `[a] = [(F)/(t)] = [MLT^(-3)]` and `[b] = [(F)/(t^(2))] = [MLT^(-4)]` |
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| 4. |
A person measures two quantities as `A = 1.0 m +- 0.2 m ,B = 2.0 m +- 0.2 m` We should report correct value for `sqrt(AB)` asA. `1.4 m +- 0.4 m`B. `1.41 m +- 0.15 m`C. `1.4 m + 0.3 m`D. `1.4 m +- 0.2 m` |
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Answer» Correct Answer - d Given `A = 1.0m +- 0.2m, B= 2.0 m +- 0.2 m` Let `V = sqrt(AB) = sqrt((1.0)(2.0)) = 1.414 m` Rounding off to two significant digit `Y = 1.4m` `(Delta Y)/(Y) = (1)/(2)[(Delta A)/(A) + (Delta B)/(B)]` `= (1)/(2)[(0.2)/(1.0) + (0.2)/(2.0)] = (0.6)/(2 xx 2.0)` `rArr Delta Y = (0.6Y)/(2 xx 2.0) = (0.6 xx 1.4)/(2 xx 2.0) = 0.20` Rounding off to one significant digit Thus currect value for `sqrt(AB) = r + Delta r = 1.4 +- 0.2m` |
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| 5. |
If the time period `(T)`of vibration of a liquid drop depends on surface tension `(S)` , radius`( r )` of the drop , and density `( rho )` of the liquid , then find the expression of `T`.A. `T = k sqrt(rhor^(3)//S)`B. `T = k sqrt(rho^(1//2)r^(3)//S)`C. `T = k sqrt(rhor^(3)//S^(1//2))`D. None of these |
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Answer» Correct Answer - a Let `T prop S^(0)t^(2) rho^(2)` by substituting the demension of `[T]= [T]` `[S] = [ML^(-2)], [r] = [L],= [ML^(-3)]` and by comparing the power of both the sides `x = -1//2, y = 3//2, z= 1//2` so`T prop sqrt(rho r^(3)//S) rArr T = k sqrt((rho r^(3))/(S))`. |
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| 6. |
In `S=a+bt+ct^2`. S is measured in metres and t in seconds. The unit of c isA. `ms^(2)`B. `m`C. `ms^(-1)`D. `ms^(-2)` |
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Answer» Correct Answer - d `ct^(2)` must have dimension of L `rArr` c must have dimensions of `L//T^(2)` i.e., `LT^(2)` |
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| 7. |
If equation `int (dt)/(sqrt(3a - 2t^(2))) = a^(x) sin ^(-1) ((r^(2))/(a^(2)) - 1)`, the value of x isA. `(3)/(2) `B. `0`C. `(1)/(2)`D. `-(1)/(2)` |
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Answer» Correct Answer - b `((t^(2))/(a^(2)) - 1)` is dimensionless ` :. [a] = [t]` As, `sqrt((3at - 2t^(2))) = [r]` `:. [(dt)/sqrt((3at - t^(2)))] = ([t])/([t])=[M^(0)L^(0)T^(0)]` `a^(2)` should be dimensionless so `x =0` |
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| 8. |
If momentum `(p)`, area `(A)` and time`(t) `are taken to be fundamental quantities then energy has the dimensional formulaA. `[p^(1)A^(-1)t^(-1)]`B. `[p^(2)A^(1)t^(1)]`C. `[p^(1)A^(1//2)t^(1)]`D. `[p^(1)A^(1//2)t^(-1)]` |
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Answer» Correct Answer - d Let enegry `E = kp^(a) A^(b)t^(c )`…(i) where is k a dimensionless constant proportionality equating dimension an both sides of(i) we get `[ML^(2)T^(-2)] = [MLT^(-1)]^(a) [M^(0)L^(2)T^(0)]^(b) [M^(0)L(0)T]^(c)` `[L]= [M^(a)L^(a+2b)T^(a+c)]` Appliying the principle of homogenety of dimensions we get `a = 1`...(ii) `a + 2b = 2`...(iii) `-a+c = -2`...(iv) On solving eqs `(ii),(iii)` and `(iv)` we get `a = 1, b = (1)/(2) , c=-1` `:. [E] = [p^(1)A^(1//2)c^(-2)]` |
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| 9. |
`int (dt)/(sqrt(2at - t^(2))) = a^(2) sin ^(-1)[[1)/(a) - 1]`The value of x isA. `1`B. `-1`C. `0`D. `2` |
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Answer» Correct Answer - c The quantity `(1)/(a) - 1` is dimensionless i.e.`[a] = [r]` `:. [sqrt(2at - t^(2)] = [r]` or, `[(dt)/(sqrt(2at -t^(2)))]= [(t)/(l)] = [M^(0) L^(0)T^(0)]` i.e. `a^(2)` should also be dimensionless or `x= 0` |
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| 10. |
Density of a liquid in CGS system is `0.625(g)/(cm^3)`. What is its magnitude is SI system?A. `0.625`B. `0.0625`C. `0.00625`D. `625` |
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Answer» Correct Answer - d CGS SI `n_(1)u_(1) = n_(2)u_(2)` `n_(1)[M_(1)L_(1)^(-3)] = n_(2) [M_(2)L_(2)^(-3)]` `:. N_(2) = n_(1)[(M_(1))/(M_(2))] xx [(1 cm)/(1 m)]^(-3)` `= 0.625 xx 10^(-3) xx 10^(6)= 625`. |
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| 11. |
The velocityof a body is given by the equation `v = (b)/(t) + ct^(2) + dt^(3)` The dimensional formula of b isA. `[M^(0)LT^(0)]`B. `[ML^(0)T^(0)]`C. `[M^(0)L^(0)T]`D. `[MLT^(-1)]` |
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Answer» Correct Answer - a In the equation left hand side has the dimension of velocity .Then the principal of homogenelty each term in the right hand side should have the dimension of velocity `[(b)/(t)] = [v]` or `[b] = [vi] =[L]` |
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| 12. |
Measure of two quantities along with the precision of respective measuring instrument is `A=2.5ms^(-1)+-0.5ms^(-1)`,B=0.10s+-0.01s`. The value of AB will beA. `(0.25 +- 0.08)m`B. `(0.25 +- 0.5)m`C. `(0.25 +- 0.05)m`D. `(0.25 +- 0.135)m` |
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Answer» Correct Answer - a `A = 2.5ms^(-1) +- 0.5ms^(-1) ,B = 0 10s` `x = AB = (2.5)(0.10) = 0.25m` `(Delta x)/(x) = (Delta A)/(A) + (Delta B)/(B)` `(0.2)/(2.5) + (0.01)/(0.10) = (0.05 + 0.025)/(0.25) = (0.075)/(0.25)` `Delta x = 0.075 = 0.08m` Rounding off to two significant figure `AB = (0.25 +- 0.08)m` |
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| 13. |
The numbers 2.745 and 2.735 on rounding off to 3 significant figures will giveA. `2.75 and 2.74`B. `2.74 and 2.73`C. `2.75 and 2.73`D. `2.74 and 2.74` |
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Answer» Correct Answer - d Rounding off `2.745` to `3` significant figure it would be `2.74` Rounding off `2.735` to `3` significant figure it would be `2.74` |
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| 14. |
Assertion: Angle and angular displacement a dimensionless quantities. Reason: Angle is equal in are length divided by radius.A. If both assertion and reason are true and reason is the correct explation of assertion.B. If both assertion 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 - a Angle or angular displacement `= ("Arc length" )/("Radius") = ([L])/([L]) .` it is a dimensionless quantity. |
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| 15. |
Assertion: Dimensional constant are the quantites whose value are constant. Reason: Dimensional constant are Dimensionless.A. If both assertion and reason are true and reason is the correct explation of assertion.B. If both assertion 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 Dimensional constants are the quantities whose value are constant and they posses dimension for example velocity of light in vacume unives gravitational constant boltamant plank constant etc. |
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| 16. |
Assertion: The dimensional formula of surface energy is `[M^(0)L^(2)T^(-2)]`. Reason: surface energy has same dimensions as that of potential energy.A. If both assertion and reason are true and reason is the correct explation of assertion.B. If both assertion 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 Dimensional formula of surface energy is `[ML^(0)T^(-2)]`. |
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| 17. |
Inductance L can be dimensional represented asA. `ML^(2)T^(2)A^(-2)`B. `ML^(2)T^(-4)A^(-3)`C. `ML^(-2)T^(2)A^(-2)`D. `ML^(2)T^(4)A^(3)` |
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Answer» Correct Answer - a `E = (1)/(2)Li^(2)` hence `L = [ML^(2)T^(-2)A^(-2)]` |
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| 18. |
Dimensional formula for latent heat is________A. `M^(0)L^(2)T^(2)`B. `MLT^(-2)`C. `ML^(2)T^(2)`D. `ML^(2)T^(1)` |
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Answer» Correct Answer - a `Q = mL rArr L = (Q)/(m)`(Heat is a energy) `= (ML^(2)T^(-2))/(M)= [M^(0)L^(2)T^(-2)]` |
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| 19. |
If `(A)/(mu_(0))` has the dimensions `[MLT^(-4)]` what is A?A. square of electric fluxB. square of magnitic fluxC. square of electric fieldD. square of energy |
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Answer» Correct Answer - c `[(A)/(mu_(0))]= [(epsilon_(0)A)/(epsilon_(0)mu_(0))]= ([epsilon_(0)d])/(1//"speed of light"]^(2)` `= [ML^(2)T^(-2)]` (Given) `[epsilon_(0) A] = [ML^(-1)T^(-2)] = ([ML^(2)T^(-2)])/([L^(2)])` So `epsilon_(0)A` is the energy per volume Thus A should be `E^(2)` |
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| 20. |
The dimensions of shear modulus areA. `MLT^(-1)`B. `ML^(2)T^(-2)`C. `ML^(-1)T^(-2)`D. `MLT^(2)` |
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Answer» Correct Answer - c Shear modulus `= ("Shearing stress")/("Shearing strain") = (F)/(Atheta) = [ML^(-1)T^(-2)]` |
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| 21. |
Assertion : The period change in time period is `1.5% ` if the length of simple pendulum increases by `3%`. Reason : Time period is dinesty proportional to length of pendulum.A. If both the asseration and reason are true and reason is a true explanation of the asseration.B. If both the asseration and reason are true but the reason is not the correct explanation of asseration.C. If the asseration is ture but reason is false.D. If both the asseration and reason are false. |
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Answer» Correct Answer - c Time period of simple pendulum of length `l` is ` T = 2 pi sqrt((I)/(g))` `rArr T prop sqrt(1)` `:. (Delta T)/(T) = (1)/(2) (Delta I)/(I) ` `(Delta T)/(T) = (1)/(2) xx 3 = 1.5%` |
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| 22. |
If energy `(E )` , velocity `(V)` and time `(T)` are chosen as the fundamental quantities , the dimensions formula of surface tension will beA. `[EV^(-2)T^(-1)]`B. `[EV^(-1)T^(-2)]`C. `[EV^(-2)T^(-2)]`D. `[E^(-2)V^(-1)T^(-3)]` |
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Answer» Correct Answer - c Let surface tension `[e] = [E^(x)V^(y)T^(z)]` we have `[x] = [MT^(-2)]` `[E] = [ML^(2)T^(-2)], [V] = [LT^(-1)] and [T] = [T]` `rArr [S] = [ML^(2)T^(-2)]^(x) [LT^(-1)]^(y) [T]^(z)` `[M^(1)L^(0)T^(-2)] = [M^(x)L^(2x +y)T^(-2x -y +z)]` `x = 1` `2x + y = 0 rArr y = -2` `-2x - y + z = -2` `rArr -2 xx 1 - (-2) + z = -2` `rArr z = -2` Hence `[s] = [EV^(-2)T^(-2)]` |
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| 23. |
If velocity , time and force were chosen as basic quantities , find the dimensions of mass and energy .A. `FVT`B. `FVT^(2)`C. `F^(0)VT^(-1)`D. `FV^(2)T^(-1)` |
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Answer» Correct Answer - a `[W] = ML^(2)T^(-2) = [F]^(a)[V]^(b) [T]^(c)` `rArr ML^(2)T^(-2) = [MLT^(-2)]^(a) [LT^(-1)]^(b) [T]^(c)` `rArr a = 1, a + b = 2 rArr b = 1` and `c - 2a - b = -2 rArr c = 1` `So [W]= FVT` |
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| 24. |
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 the asseration and reason are true and reason is a true explanation of the asseration.B. If both the asseration and reason are true but the reason is not the correct explanation of asseration.C. If the asseration is ture but reason is false.D. If both the asseration 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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| 25. |
Of the following quantities , which one has the dimensions different from the remaining three?A. Energy per unit volumeB. Force per unit areaC. Product of volume and change per unit volumeD. Angular momentum per unit mass |
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Answer» Correct Answer - d Energy per unit volume`= ([ML^(2)T^(2)])/([L^(3]])= [ML^(-3)T^(-2)]` force per unit area `= ([ML^(2)T^(2)])/([L^(2]])= [ML^(-1)T^(-2)]` Producted of voltage and charge per unit volume `= (V xx Q)/("Volume") = ("VIt")/("Volume")= ("Power" xx "time")/("Volume")` `rArr ([ML^(2)T^(-3)][T])/([L^(3)] = [ML^(-1)T^(-2)]` Angular momentum per unit mass `([ML^(2)T^(-1)])/([M]) = [L^(2)T^(-2)]` So angular per unit has different dimension |
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| 26. |
To estimate `g` (from `g = 4 pi^(2)(L)/(T^(2))`), error in measurement of `L` is `+- 2%` and error in measurement of `Tis +- 3%` The error in estimated `g` will beA. `+- 8%`B. `+- 6%`C. `+- 3%`D. `+- 5%` |
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Answer» Correct Answer - a `g = 4 pi^(2) (L)/(T^(2))` `(Delta L)/(L) = 2% = +- 2 xx 10^(-2)` `(Delta T)/(T) = 3% = +- 3 xx 10^(-2)` `rArr (Delta g)/(g) = (Delta L)/(L) +(2Delta T)/(T) =2 xx 10^(-2)+ 2 xx 3 xx 10^(-2)` `8 xx 10^(-2)= +- 8%` |
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| 27. |
A stone lying at rest in a river The minimum mass of stone, `m = krhov^(x) g^(-3)` is needed for remaining at rest here, `h =` constant having no unit , `g =` acceleration due to gravity , `v =` river flor velocity , `rho =` density of water. The value of x isA. `3`B. `5`C. `6`D. `8` |
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Answer» Correct Answer - c The unit of LHS is `kg` The unit of RHS should be `kg` Here `krhov^(2)g^(-3)= ((kg)/(m^(3))) ((m)/(s))^(x)((m)/(s^(2)))^(-3)` `= (kg) (m)^(-3+x - 3)(s)^(+6-x)` `:. -6 +x = 0` `rArr x = 6` |
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| 28. |
Drift speed of electron inside the metallic conductor is `v_(A) = eE^(y) m^(z) tau` (here, `e =` electronic charge, `E =` electric field, `m =` mass of electron and `tau =` time relaxation). Find the value of `y`.A. `(3)/(2) `B. `0`C. `(1)/(2)`D. `1` |
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Answer» Correct Answer - d The unit of LHS `= m^(1)s^(-1)` The unit of RHS `="coulomb"(("newton")/("colomb"))^(1) ("kilogram")^(2)("second")` But The unit of LHS= unit of RHS Colomb does not come in RHS `(("colomb"^(1))/("colomb")^(2)) =` unitless `rArr y = 1` |
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| 29. |
Position of a body with acceleration `a` is given by `x=Ka^mt^n`, here t is time Find demension of m and n.A. `m = 1, n = 1`B. `m = 1, n = 2`C. `m = 2, n = 1`D. `m = 2, n = 2` |
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Answer» Correct Answer - b As `x = ka^(m)xx t^(n)` `[M^(0)L^(1)T^(0)] = [LT^(-2)]^(m)[T^(n)]= L^(m)T^(-2m+n)` `:. M = 1` and `-2m + n = 00 rArr n = 2` |
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| 30. |
The relation `tan theta = v^(2)// r g` gives the angle of banking of the cyclist going round the curve . Here `v` is the speed of the cyclist , `r` is the radius of the curve , and `g` is the acceleration due to gravity . Which of the following statements about the relation is true ?A. both dimensionally and numerically correctB. neithen numerically not dimensionally correctC. dimensionally correct onlyD. numerically correct only |
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Answer» Correct Answer - c Given equation is dimension correct because both sides are dimensionaless but numerically wrong because the correct equation is `tan theta = (v^(2))/(rg)` |
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| 31. |
A physical quantity `x` depends on quantities `y and z` as follows : ` x = Ay + B tan ( C z)`, where `A , B and C` are constants. Which of the followings do not have the same dimensions?A. x and BB. `C and z^(-1)`C. y and B//A`D. x and A |
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Answer» Correct Answer - d `x = Ay + B tan Cz` From the dimensional homogenetly `[x] - [Ay] = [B] rArr ([x)/(A]) = [y] = ([B)/(A))]` `[cz] = [M^(0) L^(0)T^(0)] =` Dimension less `x` and `A` have the different dimensionsx and `B, C` and `z^(-1), y` and `(B)/(A)` have the same dimension but `x` amd `A` have the different dimensions. |
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| 32. |
In dimension of circal velocity `v_(0)` liquid following through a take are expressed as `(eta^(x) rho^(y) r^(z))` where `eta, rhoand r `are the coefficient of viscosity of liquid density of liquid and radius of the tube respectively then the value of `x,y` and `z` are given byA. `1,1,1`B. `1,-1,-1`C. `-1,-1,1`D. `-1,-1,-1` |
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Answer» Correct Answer - b `v_(c) = [eta^(x) rho^(y)r^(z)]` `[L^(1)T^(-1)] = [M^(1)L^(-1) T^(-1)]^(x) [M^(1)L^(-3)]^(y) [L^(1)]^(z)` `[L^(1)T^(-1)] prop [M^(x+y)] [L^(-x +3y +z)] [T^(-x)]` taking comparison an both sides `x + y = 0 , -x - 3y + z= 1 , -x = -1` `rArr x=1 , y= -1 , z= -1` |
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| 33. |
The magnetic moment has dimensions ofA. `[LA]`B. `[L^(2)LA]`C. `[LT^(-1)A]`D. `[L^(3)T^(-1)A]` |
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Answer» Correct Answer - b The magnetic moment of a current carriying loop is define as the product of current in the loop with the area of loop in vector from , `vec M = 1 vec A` Thus dimesion of `[M] = [A] [L^(2)] = [L^(2)A] ` |
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| 34. |
In a circuit potential difference across resistence `V = (4 +- 0.25)V `and curent in resistence ,`f = (1 +- 0.1)` what is the value of resistence with its percentage errorA. `(4 +- 0.4)Omega`B. `4 Omega + 16.25%`C. `4 Omega + 18.25%`D. `4 Omega + 22.25%` |
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Answer» Correct Answer - b `E_(0) = (V_(0))/(I_(0)) = (4)/(1) = 4 Omega` `(DeltaR)/(R ) = (Delta V)/(V) + (Delta I)/(I)` `= (0.25)/(4) + (0.1)/(1.0) = 0.1625` `R = 4Omega + 16.25%` |
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| 35. |
The period of oscillation of a simple pendulum is given by `T=2pisqrt((l)/(g))` where l is about 100 cm and is known to have 1 mm accuracy. The period is about 2 s. The time of 100 oscillation is measrued by a stop watch of least count 0.1 s. The percentage error is g isA. `0.1%`B. `1%`C. `0.2%`D. 0.8%` |
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Answer» Correct Answer - c `T = 2 pi sqrt(1//g)rArr T^(2) = 4 pi^(2) 1//g rArr g = (4 pi^(2)l)/(T^(2))` `Here % error in l = (1 mm)/(100cm) xx 100 = (0.1)/(100) xx 100 = 0.1%` and `% "error in" T = (0.1)/(2 xx 100) xx 100 = 0.05%` `:. % "error in" g = % "error in" l + 2(% "error in" T)`. |
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| 36. |
The mean length of an object is 5 cm. Which of the following measurements is most accurate?A. `4.9 cm`B. `4.805 cm`C. `5.25 cm`D. `5.4 cm` |
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Answer» Correct Answer - a Given length `l= 5 cm` Now , chacking the error with each option one by one we get `Delta l_(1) = 5 - 4.9 = 0.1cm` `Delta l_(2) = 5 - 4.805 = 0.195 cm` `Delta l_(3) = 5.25 - 5 = 0.25 cm` `Delta l_(4) = 5.4 - 5 = 0.4 cm` Error `Delta l_(1)` is least, Hence `4.9 cm` is most precise |
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| 37. |
If force `(F)` velocity `(V)` and time `(T)` are taken as fundamental units, then the dimensions of mass areA. `[FVT^(-1)]`B. `[FVT^(-2)]`C. `[FV^(-1)T^(-1)]`D. `[FV^(-1)T]` |
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Answer» Correct Answer - d Let `m prop F^(x)V^(y)T^(z)` By substituting the following dimension `[F] = [MLT^(-2)],[V] = [LT^(-1)],[T]=[T]` `M^(1)L^(0)T^(0)] = [MLT^(-1) [LT^(-1)][T]` `rArr [M^(1)L^(0)T^(0)] = [M^(x)L^(x + y) T^(-2x -y +z)]` `x = 1` `x + y =0 , rArr y = -1` ` rArr -2 xx 1 - (-1) + z = 0` `rArr z = 1` Hence `[M] = [F^(1)V^(-1)T^(1)] = [FV^(1)T]` |
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| 38. |
Which of the following following is not equal to watt?A. `"Joule" // "second"`B. `"Ampere" xx "volt"`C. `("Ampere")^(2) xx "ohm"`D. `"Ampere" // "volt"` |
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Answer» Correct Answer - d `"Watt" = "Joule"// "second" = "Ampere" xx "volt" = "Ampere"^(2) xx "Ohm"` |
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| 39. |
The velocity of a freely falling body changes as `g^ph^q`where g is acceleration due to gravity and h is the height. The values of p and q areA. `1,(1)/(2)`B. `(1)/(2),(1)/(2)`C. `(1)/(2),1`D. `1,1` |
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Answer» Correct Answer - b `x prop g^(p)h6^(q)` (given) By substituting the dimension of each quantity and comparing the powers in both sides we get `[LT^(-1)] = [LT^(-2)]^(p)[L]^(q)` `rArr p + q = 1, 2p = -1, :. P = (1)/(2), q= (1)/(2)` |
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| 40. |
If the acceleration due to gravity is represented by unity in a system of unit and one second is the unit of time , the unit length isA. `9.8 m`B. `1 m`C. `98 m`D. `0.98 m` |
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Answer» Correct Answer - a `1 Ls^(-2)= 9.8 ms^(-2)` or,`l = 9.8m` |
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| 41. |
Newton - second is the unit ofA. VelocityB. Anguler momentumC. MomentumD. Energy |
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Answer» Correct Answer - c Impulse = change in momentum `= Fxx t` So the unit of momentuim will be equal to `"Newton" -sec`. |
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| 42. |
Which of the following following is not a unit of energy?A. `W- s`B. `kg-m//sec`C. `N-m`D. Joule |
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Answer» Correct Answer - b `kg-m//sec` is the unit of linear momentum |
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| 43. |
A suitable unit for gravitional constant isA. `kg m sec^(-1)`B. `Nm^(-1) sec`C. `N m^(2) kg^(-2)`D. `kg m sec` |
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Answer» Correct Answer - c `F= (Gm_(1)m_(2))/(d^(2)), :. G = (Fd^(2))/(m_(1)m_(2)) = Nm^(2)//kg^(2)` |
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| 44. |
The unit of potential energy isA. `g(cm//sec^(2))`B. `g(cm//sec)^(2)`C. `g(cm^(2)//sec)`D. `g(cm//sec)` |
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Answer» Correct Answer - b Potential energy `= mgh = g((cm)/(sec^(2))) cm= g((cm)/(sec))^(2)`. |
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| 45. |
Which of the following following is not the unit of energy?A. CalorieB. JouleC. Electron voltD. Watt |
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Answer» Correct Answer - d What is a unit of power. |
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| 46. |
Which of the following following is smallest unit?A. MillimeterB. AngstromC. FermiD. Metre |
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Answer» Correct Answer - c `1 "ferm" = 10^(-15) "metre"` |
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| 47. |
A watt isA. `kg m//s^(2)`B. `kg m^(2)//s^(2)`C. `kg m//s`D. `kg m^(2)//s^(2)` |
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Answer» Correct Answer - b `Watt= (J)/(s) = (N - m)/(s) = kg(m^(2))/(s^(3))` |
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| 48. |
In the formula `X = 3YZ^(2),X and Z` have dimensions of capacitance and magnetic induction respectively. The dimensions of `Y` in MKSQ system are ………………, ………………….A. `[M^(2)L^(-2)T^(3)Q^(4)]`B. `[M^(-3)L^(-2)T^(4)Q^(4)]`C. `[M^(-2)L^(-2)T^(4)Q^(3)]`D. `[M^(-3)L^(-1)T^(3)Q^(4)]` |
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Answer» Correct Answer - b `C = (q)/(V) = (q^(2))/(W)` `[X] rarr [C] = [M^(-1)L^(-2)T^(2)Q^(2)` And `[B] = ([F)/(iL])` `[Z] rarr [B] = [(MLT^(-2))/(QT^(-1)L)]= [MT^(-1)Q^(-1)]` given `Y = (X)/(3Z^(2))` `[T] =[(M^(-1)L^(-2)T^(2)Q^(2)])/(MT^(-1)Q^(-1))]^(2) = [M^(-3)L^(-2)T^(-4)Q^(4)]`. |
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| 49. |
Assertion: The given equation `x = x_(0) + u_(0)t + (1)/(2) at^(2)` is dimensionsally correct, where x is the distance travelled by a particle in time t , initial position `x_(0) ` initial velocity `u_(0)` and uniform acceleration a is along the direction of motion. Reason: Dimensional analysis can be used for cheking the dimensional consistency or homogenetly of the equation.A. If both assertion and reason are true and reason is the correct explation of assertion.B. If both assertion 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 - a Given equation `x = s_(0)+ u_(0)t + (1)/(2) at^(2)` The dimensions of each term mey be written as `[x] = [L]` `[x_(0)] = [L]` `[u_(0)t]= [LT^(-1)][T]= [L]` `[((1)/(2)) at^(2)] = [LT^(-2)][T^(2)] = [L]` As each term on the right hand side of the equation the same dimension as left hand side of the equal hence this equation id dimensionally correct. |
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| 50. |
If `u_(1) and u_(2)` are the selected in two system of measurement and `n_(1) and n_(2)` their nomerical values, thenA. `n_(1)u_(1)=n_(2)u_(2)`B. `n_(1)u_(1)+n_(2)u_(2) = 0`C. `n_(1)n_(2)=u_(1)u_(2)`D. `(n_(1) + u_(1)) =(n_(2) + u_(2))` |
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Answer» Correct Answer - a Physical quantity `(p) = "Numerical value" (n) xx "Unit" (u)`. if physical quantity remain constant then `n = 1//u :. N_(1)u_(1) = n_(2)u_(2)` |
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