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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. |
Using the expression `2d sin theta = lambda`, one calculates the values of `d` by measuring the corresponding angles `theta` in the range `0 to [email protected]`. The wavelength `lambda` is exactly known and error in `theta` is constant for all values of `theta`. As `theta` increases from `[email protected]`A. the absolute error in `d` remains constant.B. the absolute error in `d` increases.C. the fractional error in `d` remain constant.D. the fractional error in `d` decreases. |
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Answer» Correct Answer - D `d = ( lambda)/( 2 sin theta) In d = In ((lambda)/( 2)) - In sin theta ( Delta d)/( d) = 0 - ( cos theta d theta)/( sin theta)` `(( Delta d)/(d))_(max) = +- cot theta Delta theta` ` ( lambda)/( 2 sin theta) cot theta Delta theta = ( lambda)/(2) ( cos theta)/( sin^(2) theta) Delta theta` As `theta` increases , `cot theta` decreases and `(cos theta)/( sin^(2) theta)` also decreases. |
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| 2. |
Using the expression `2d sin theta = lambda`, one calculates the values of `d` by measuring the corresponding angles `theta` in the range `0 to [email protected]`. The wavelength `lambda` is exactly known and error in `theta` is constant for all values of `theta`. As `theta` increases from `[email protected]`A. the absolute error in `d` remains constant.B. the absolute error in `d` increases.C. the fractional error in `d` remain constant.D. the fractional error in `d` decreases. |
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Answer» Correct Answer - D `d = ( lambda)/( 2 sin theta) In d = In ((lambda)/( 2)) - In sin theta ( Delta d)/( d) = 0 - ( cos theta d theta)/( sin theta)` `(( Delta d)/(d))_(max) = +- cot theta Delta theta` ` ( lambda)/( 2 sin theta) cot theta Delta theta = ( lambda)/(2) ( cos theta)/( sin^(2) theta) Delta theta` As `theta` increases , `cot theta` decreases and `(cos theta)/( sin^(2) theta)` also decreases. |
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| 3. |
A wire of length `l = +- 0.06cm` and radius `r =0.5+-0.005 cm` and mass `m = +-0.003gm`. Maximum percentage error in density isA. `4`B. `2`C. `1`D. `6.8` |
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Answer» Correct Answer - A `rho = (m)/( pi r^(2) l)` `rArr ((Delta rho)/( rho) xx 100) = ((Delta m)/( m) + ( 2 Delta r)/( r ) + ( Delta l)/( l)) xx 100` ` = ((0.003)/( 0.3) + ( 2 xx 0.005)/( 0.5) + ( 0.06)/( 6)) xx 100 = 4%` |
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| 4. |
If `x and a` stand for distance , then for what value of `n` is the given equation dimensionally correct? The equation is `int (dx)/( sqrt ( a^(2) - x^(n))) = sin^(-1) (x)/(a)`A. `0`B. `2`C. `-2`D. `1` |
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Answer» Correct Answer - B Here `x^(n)` has the same dimensions as `a^(2)` . Thus , `n = 2` will make the expression dimensionally correct. |
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| 5. |
Test dimensionally if the `v^2=u^2+2ax` may be correct. |
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Answer» We have `v^(2) - u^(2) = 2as`. Checking the dimensions on both sides , we get `LHS = [ LT^(-2)]^(2) - [LT^(-1)]^(2) = [L^(2)T^(-2)] - [L^(2)T^(-2) ] - = [L^(2) T^(-2)]` `RHS = [L^(1)T^(2)] [L] = [L^(2) T^(2)]` Comparing `LHS and RHS` , we find `LHS = RHS`. Hence, the formula is dimensionally correct. |
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| 6. |
Check whether the relation `S = ut + (1)/(2) at^(2)` is dimensionally correct or not , where symbols have their usual meaning . |
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Answer» We have `S = ut + (1)/(2) at^(2)` Checking the dimensions on both sides, `LHS = [M^(0) L^(1) T^(0)]` `RHS = [LT^(-1)] [ T] + [LT^(-2)] [ T^(2)]` `= [M^(0) L^(1) T^(0) ] + [ M^(0) L^(1) T^(0) ] = [ M^(0) L^(1) T^(0)]` Comparing the `LHS and RHS` , we get `LHS = RHS` Hence the formula is dimensionally correct. |
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| 7. |
If `E , M , J , and G` , respectively , denote energy , mass , angular momentum , and gravitational constant , then `EJ^(2) //M^(5) G^(2)` has the dimensions ofA. TimeB. AngleC. MassD. Length |
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Answer» Correct Answer - B ` (E J ^(2))/(M^(5) G^(2)) = ([ML^(2)T^(-2)][ML^(2)T^(-1)]^(2))/(M^(5)xx[M^(-1)L^(3)T^(-2)]^(2)) = M^(0) L^(0) T^(0)` This comes out to be dimensionless and angle is also dimensionless. |
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| 8. |
If `L and R` denote inductance and resistance , respectively , then the dimensions of `L//R` areA. ` M^(1) L^(0) T^(0) Q^(-1)`B. ` M^(0) L^(0) T Q^(0)`C. ` M^(0) L^(1) T^(-1) Q^(0)`D. ` M^(-1) L T^(0) Q^(-1)` |
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Answer» Correct Answer - B Dimension of `L//R` is same as that of time. |
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| 9. |
Find the dimensions of capacitance.A. `[M^(-1) L^(-2) T A^(2)]`B. `[M^(-1) L^(-2) T^(3) A^(2)]`C. `[M^(-1) L^(-2) T^(4) A^(2)]`D. `[M^(-1) L^(-2) T^(2) A^(2)]` |
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Answer» Correct Answer - C Capactiance , `C = ("Charge")/("Potential") = (AT) /( ML^(2)T^(-3)A^(-1)) = [ M^(-1) L^(-2) T^(4)A^(2)]` |
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| 10. |
Which of the following pairs have the same dimensions ? (L = inductance , C = capacitance , R = resistance)A. `(L)/( R ) and CR`B. ` LR and CR`C. `(L)/( R) and sqrt(L C)`D. `R C and (1)/( L C )` |
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Answer» Correct Answer - A::C `L//R , CR` and `sqrt(LC )` all have dimensions of time `[T]`. |
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| 11. |
The SI and CGS units of energy are joule and erg respectivel. How many ergs are equal to one joule. |
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Answer» Joule: SI system , erg: CGS system Work = Force `xx` Distance = Mass `xx` Acceleration `xx` Length = Mass `xx ( "length")/("Time")^(2) xx "Length"` Dimensions of work =` [W] = [ M^(1) L^(2) T^(2) ]` `:. a = 1, b = 2 , c = -2` Now `|{:(SI" system",,CGS" system"),(M_(1)=1kg,,M_(2)=1g),(L_(1)=1m,,L_(2)=1cm),(T_(1)=1s,,T_(2)=1s):}|` Here `N_(1) = 1 , N_(2) = ?` `:.` Using `N_(2) = N_(1) [(M_(1))/(M_(2))]^(a) [[(L_(1))/(L_(2))]^(b)[(T_(1))/(T_(2))]^(c)` `= 1[( 1 kg)/( 1 g)]^(1) [( 1 m)/( 1cm)]^(1) [( 1 s)/( 1 s)]^(-2) = 10^(5)` `1 [ (1000 g)/( 1 g)]^(1) [ (100 cm)/(1 cm)]^(2) = 10^(7)` ` :.N_(2) = 10^(7)` So `1 J = 10^(7) erg` |
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| 12. |
To find the distance `d` over which a signal can be seen clearly in foggy conditions, a railways-engineer uses dimensions and assumes that the distance depends on the mass density `rho` of the fog, intensity (power/area) `S` of the light from the signal and its frequency `f`. the engineer finds that `d` is proportional to `S^(1//n)`. the value of `n` is |
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Answer» Correct Answer - `3` `d = rho^(a) S^(b) f^( c)` ` d = ((kg)/( m^(3)))^(a) xx (( w)/(m^(2))) ^(b) xx ((1)/( sec))^(c )` `d = (ML^(-3))^(a) ((ML^(2) T^(-3))/(L^(2)))^(b) ((1)/( sec))^( c )` `L^(1) = M^(a + b L^(-3a) T^(-3b -c )` ` -3a = 1 rArr a = -(1)/(3)` ` a + b = 0` `( :. b = (1)/(3))` `- 3b -c = 0` `( :. c = 1)` `:. rho^-(1)/(3) S^(1)/(3) f^(1)` `:. d prop S^(1//n) ( :. n = 3)` |
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| 13. |
The dimensions of electrical conductivity are …….. |
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Answer» `sigma = (1)/( rho) = (l)/(R A) = (l)/((V)/(I)A) = (Il)/(VA) = (Il)/((W)/(q) A) = (I l q)/(WA) xx (t)/(t)` `= (I^(2)"lt")/( WA) = (A^(2) LT^(1))/( ML^(2) T^(-2) L^(2)) = [M^(-1) L^(-3) T^(3) A^(2)]` |
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| 14. |
Convert gravitational constant (G) from CGS t o MKS system. |
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Answer» The dimensional formula of G is `[M^(-1) L^(3) T^(-2) ]`, i.,e `a = -1 , b = 3 , c = -2`. By using `n_(2) = n_(1) [(M_(1))/(M_(2))]^(a) [[(L_(1))/(L_(2))]^(b)[(T_(1))/(T_(2))]^(c)` Value of G in CGS unit `= 6.67 xx 10^(-8) cm^(3) g s^(-2)` `= 6.67 xx 10^(-8) [(g)/(kg)]^(-1) [(cm)/(m)]^(3) [(s)/(s)]^(-2)` `= 6.67 xx 10^(-8) [ (g)/(10^(3) g)]^(-1) [ ( cm)/( 10^(2) cm)]^(3) [ (s)/(s)]^(-2)` = 6.67 xx 10^(-11)` `:. = 6.67 xx 10^(-11)` MKS units |
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| 15. |
A force `F` is given by `F = at + bt^(2)` , where `t` is time . What are the dimensions of `a and b`? |
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Answer» From the principle of dimensional homogenity , `[ F] = [ at]` :. ` [ a] = [(F)/(t) ] = [ (MLT^(-2))/( T)] = [ MLT^(-3)` Similarly ,`[F] = [ bt^(2)]` :. `[b] = [(F)/(t^(2))] = [ (MLT^(-2))/( T^(2))] = [ MLT^(-4)]` |
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| 16. |
In `CGS` system the magnitude of the force is 100 dynes. In another system where the fundamental phyical quamtities are kilogram , meter , and minute, find the magnitude of the force. |
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Answer» We have `n_(1) = 100 , M_(1) = g , L_(1) = cm , T_(1) = s and M_(2) = kg , L_(2) = m ,T_(2) = min`, the dimensional formula of forces is `[M^(1) L^(1) T^(2)]`. Where `a = 1 , b = 1 , c = -2`. By substituting these values in the following conversion formula, we have `n_(2) = n_(1) [(M_(1))/(M_(2))]^(a) [[(L_(1))/(L_(2))]^(b)[(T_(1))/(T_(2))]^(c)` ` = 100 [ (g)/(kg)]^(1) [ (cm)/( m)]^(1) [( s)/( min)]^(-2)` `= 100[(g)/(10^(3)g)]^(1) [(cm)/(10^(2) cm)]^(1) [ (s)/( 60 s)]^(-2) = 3.6` |
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| 17. |
Given that `Y = a sin omegat + bt + ct^(2) cos omegat`. The unit of `abc` is same as that ofA. `y`B. `y//t`C. `( y //t)^(2)`D. `(y//t)^(3)` |
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Answer» Correct Answer - D ` y = a sin omega t + bt + ct^(2) cos omega t` Here `a = y , b = y//t , c = y //t^(2)` :. `a xx bxx c = y xx y//t xx y//t^(2) = ( y //t)^(3)` |
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| 18. |
The value of acceleration due to gravity is ` 980 cm s^(-2)`. What will be its value if the unit of length is kilometer and that of time is minute? |
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Answer» Correct Answer - `3.53` Dimension of acceleration due to gravity is `[LT^(-2)]`. In CGS system , Let `L_(1)` and `T_(1)` represent length and time measured in cm and second , respectively . Let `n_(2)` be the value of acceleration due to gravity in the new system . The length `L_(2)` and time `T_(2)` are measured in kilometer and minute, respectively. Now `n_(1) [ L_(1) T_(1)^(-2)] = n_(2) [ L_(2) T_(2)^(-2)]` or ` n_(2) = n_(1) [(L_(1))/(L_(2))] [(T_(1))/(T_(2))]^(-2)` `= 980 [(1)/( 10^(5))] [(1)/( 60)]^(-2) = ( 980 xx 60 xx 60)/( 10^(5)) = 3.53` |
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| 19. |
If ` x= at + bt^(2)` , where `x` is the distance rtravelled by the body in kilometer while `t` is the time in seconds , then find the units of `b`. |
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Answer» From the principle of dimensional homogenity , `[ x] = [ bt^(2)]` rArr `[b] = [(x)/( t^(2))]` Therefore , unit of `b = km s^(-2)`. |
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| 20. |
The dimensional formula for electric potential isA. `[M L^(2) T^(-3) A^(-1)]`B. `[M L T^(-3) A^(-1)]`C. `[M L^(2) T^(-3) K^(-1)]`D. none of these |
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Answer» Correct Answer - A Electric potential `= ("Work")/("Charge")= (ML^(2)T^(-2))/(AT) = [ ML^(2)T^(-3)A^(-1)]` |
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| 21. |
The effictive length of a simple pendulum is the sum of the following three : length of string , radius of bob , and length of hook. In a simple pendulum experiment , the length of the string , as measured by a meter scale , is `92.0 cm`. The radius of the bob combined with the length of the hook , as measured by a vernier callipers , is `2.15 cm`. The effictive length of the pendulum isA. `94.1 cm`B. `94.2 cm`C. `94.15 cm`D. `94 cm` |
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Answer» Correct Answer - B `( 92.0 + 2.15) cm = 94.15 cm`. Rounding off to first decimal place , we get `94.2 cm`. |
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| 22. |
A physical parameter `a` can be determined by measuring the parameters `b, c, d, and e` using the relation `a=b^(alpha)c^(beta)//d^(gamma)e^(delta)`. If the maximum errors in the measurement of `b, c , d, and e are b_(1) %,c_(1)% ,d_(1)% , and e_(1) %`, then the maximum error in the value of `a` determined by the experminent. |
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Answer» ` a = b^(alpha) c ^(beta)//d^(gamma) e^(delta)` So maximum error in `a` is given by `(( Delta a)/( a) xx 100)_(max) = alpha . (Delta b)/(b) xx 100 + beta . ( Delta c)/( c ) xx 100 + gamma . ( Delta d)/( d) xx 100 + delta . ( Delta e)/( e) xx 100 = ( alpha b_(1) + beta c_(1) + gamma d_(1) + delta e_(1))%` |
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| 23. |
If the value of resistance is `10.845 Omega` and the value of current is `3.23 A` , the potential difference is `35.02935 V`. Find its value in significant number. |
| Answer» Value of current `(3.23 A) ` has minimum significant figure `(3)`. So the value of potential difference `V ( = IR )` has only three significant figures. Hence , its value is `35.0 V`. | |
| 24. |
Which of the following numbers has least number of significant figures?A. `0.80760`B. `0.80200`C. `0.08076`D. `80.267` |
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Answer» Correct Answer - C `0.08076` has least number of significant figures , i.e., `4`. |
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| 25. |
In an experiment for determining the value of acceleration due to gravity `(g)` using a simpe pendulum , the following observations were recorded: Length of the string `(l) = 98.0 cm` Diameter of the bob `(d) = 2.56 cm` Time for `10 oscillations (T) = 20.0 s` Calculate the value of `g` with maximum permissible absolute error and the percentage relative error. |
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Answer» Correct Answer - ` ( 9.80 +- 0.11) ms^(-2)` Time period for a simple pendulum is `T = 2 pi sqrt((l_(eff))/(g))` …..(i) where `l_(eff)` is the efective length of the pendulum equal to `(l + (d)/(2))` and time period equals `T = ( 20.0)/( 10) = 2.00 s` From (i) , we get `g = (4 pi^(2) (_(eff)))/(T^(2))` To calculate actual value of `g`, Since `g = ( 4 pi^(2) (98 + 1.28))/((2.00)^(2) = 980 cm s^(-2) = 9.80 ms^(-2)` Error int he value of `g`: `(Delta g)/( g) = (Delta l_(eff))/(l_(eff)) + 2 ((Delta T)/( T)) = (Delta l + Delta r )/( l + r) + 2 ((Delta T)/(T))` Further , since errors can never exceed the least count of he measuring instrument. So , `Delta l = 0.1 cm and Delta r = 0.01 cm`. `(Delta g)/( g) = (( 0.1 + 0.01)/(98.0 + 1.28)) + 2 ((0.1)/( 20.0))` `rArr = 0.0011 + 0.01 = 0.0111` Thus , percentage error `(Delta g)/( g) xx 100 % = 1.1%` and absolute error `= Delta g = g (0.011) = 0.11 ms^(-2)` So , `g = ( 9.80 ms^(-2) +- 1.1 %) = ( 9.80 +- 0.11) m s^(-2)` |
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| 26. |
It has been observed that velocity of ripple waves produced in water `( rho)` , and surface tension `(T)` . Prove that ` V^(2) prop T// lambda rho`. |
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Answer» Correct Answer - `v^(2) prop (T)/ (lambda rho)` According to the problem , ` v prop lambda ^(a) rho^(b) T^(c )` ` v = k lambda ^(a) rho^(b) T^(c )` Where `k` is a dimensionless constant . ` LT%(-1) = L^(a) (ML^(-3))^(b) ( MT^(-2))^( c )` rArr `M^(0) L^(1) T^(-1) = M^( b + c) L^( a - 3b ) T^(-2 c )` Using the principle of homogenity , we get ` b + c = 0 , a - 3 b = 1 , -2c = -1 ` Solving these equations , we get `a = -(1)/(2) , b = -(1)/( 2) , c = (1)/(2)` So , the relation becomes ` v = k lambda ^(-1//2) rho^(-1//2) T^(1//2) rArr v^(2) prop (T)/( lambda rho)` |
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| 27. |
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`. |
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Answer» Let `T prop S^(x) r^(y) rho ^(z) or T = K S^(x) r^(y) rho^(z)` By substituting the dimension of each quantity in both sides, `[M^(0) L^(0) T^(1)] = K [MT^(-2)]^(x) [L]^(y) [ML^(-3)]^(z) = [M^(x+ z) L^( y - 3z) T^(-2x)]` By equating the power of M, L , and T in both sides, By solving above three equations , we get `x = -1//2 , y = 3//2 , z = 1//2` So the time period can be given as `T = KS^(-1//2) r^(3//2) rho^(1//2) = K sqrt( rho r^(3)/( S))` |
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| 28. |
A physical quantity depends upon five factors , all of which have dimensions, then method of dimensional analysisA. Can be appliedB. Cannot be appliedC. Depends upon factors involvedD. Both (a) and ( c)` |
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Answer» Correct Answer - B If a quantity depends upon more than three factors, each having dimensions , then the method of dimensional analysis cannot be applied. It is because applying the principle of homogenity will give only three equations. |
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| 29. |
Assuming that the mass `m` of the largest stone that can be moved by a flowing river depends upon the velocity `v` of the water , its density `rho` , and the acceleration due to gravity `g` . Then `m` is directly proportinal toA. `v^(3)`B. `v^(4)`C. `v^(5)`D. `v^(6)` |
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Answer» Correct Answer - D ` m prop v^(a) rho^(p) g^(c ) ML^(0) T^(0) prop (LT^(-1))^(a) (ML^(-3))^(b) (LT^(-2))^( c )` Comparing the powers of `M , L , and T` and solving , we get ` b = 1 , c = -3 , a = 6 rArr m prop v^(6)` |
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| 30. |
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 B`B. `C and z^(-1)`C. `y and B//A`D. ` x and A` |
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Answer» Correct Answer - D `[x] = [Ay] = [B] rArr [y] = [ B//A]` Also , `[x] != [A] and [Cz] =` dimensionless `rArr [C] = [z^(-1)]` |
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| 31. |
The energy of a system as a function of time `t` is given as `E(t) = A^(2)exp(-alphat)`, `alpha = 0.2 s^(-1)`. The measurement of `A` has an error of `1.25%`. If the error In the measurement of time is `1.50%`, the percentage error in the value of `E(t)` at t = 5 s` is |
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Answer» Correct Answer - `4` `E(t) = A^(2) e^(- alpha t)` In `E = 2 In A - alpha t` `:. (Delta E)/(E) xx 100 = 2 (Delta A)/( A) xx 100 - alpha t xx 100` (i) For time : `(Delta t)/(t) xx 100 = 1.5` At `t = 5 s, Delta t = ( 1.5 xx 5 )/( 100) = ( 7.5)/( 100)` Using Eq(i) % error in `E = 2 xx ( 1.25) + alpha xx ( 7.5)/(100) xx 100` `= 2.5 + 1.5 = 4` |
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| 32. |
A physical quantity `x` is calculated from the relation `x = ( a^(2) b^(3))/(c sqrt( d))`. If the percentage error in `a, b , c , and d are 2% , 1% , 3%, and 4%`, respectively , what is the percentage error in `x`? |
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Answer» As `x = ( a^(2) b^(3))/(c sqrt( d))`, `:. (Delta x)/(x) = +- [ 2 ( Delta a )/a) + 3 ( Delta b)/(b) + (Delta c )/( c) + (1)/(2) (Delta d)/( d) ] (Delta x)/( x) xx 100` `= +- [ 2 xx 2% + 3 xx 1% + 3% + (1)/(2 )xx 4%]` `= +- [ 4% + 3% + 3% + 2% ] = +- 12%`. |
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| 33. |
A physical quantity `P` is given by `P = (A^(3) B^(1//2))/( C^(-4) D^(3//2))`. Which quantity among `A, B , C, and D` brings in the maximum percentage error in `P`? |
| Answer» Quantity `C` has maximum power . So it brings maximum error in`P`. | |
| 34. |
A quantity` X` is given by `epsilon_(p) L(delta V)/(delta t)` , where `epsilon_(p)` is the permitivity of free space ,L is a length , `delta V` is a potential difference and ` deltat ` is a time interval . The dimensional formula for `X` is the same as that ofA. ResistanceB. ChargeC. VoltageD. Current |
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Answer» Correct Answer - D Dimensionally `epsilon_(0)L = C`, where C = capacitance Dimensionally `C Delta V = q` , where `q` is charge Dimensionally `(q)/(Delta t) = I` , where `I` is current |
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| 35. |
The dimension of `((1)/(2))epsilon_(0)E^(2)` (`epsilon_(0)` : permittivity of free space, E electric fieldA. `MLT^(-1)`B. `ML^(2)T^(-2)`C. `ML^(-1)T^(-2)`D. `ML^(2)T^(-1)` |
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Answer» Correct Answer - C Here , `(1//2) epsilon_(0) E_(2)` represents energy per unit volume . `|epsilon_(0)| [E^(2)] = ("Energy")/("Volume") = ([ML^(2)T^(-2)])/([L^(3)]) = ML^(-1)T^(-2)` |
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| 36. |
The equation of state for real gas is given by `((p + (a)/(V^(2))(V - b) = RT`. The dimension of the constant `a` is ………………. . |
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Answer» In the equation , `(P + (a)/(V^(2))) ( V - b ) = RT [P] = [(a)/( V^(2))]` or ` [a] = [PV^(2)] = [(MLT^(-2))/(L^(2)) L^(6)] = [ML^(5) T^(-2)]` |
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| 37. |
`( alpha)/( t^(2)) = Fv + ( beta)/(x^(2))` Find the dimension formula for `[ alpha] and [ beta]` ( here t = time , F = force , v = velocity , x = distance). |
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Answer» Correct Answer - `[beta] = M^(1) L^(4) T^(-3) ; [alpha] = M^(1) L^(2) T^(-1)` Since `[Fv] = M^(1) L^(2) T^(-3)`, So `[ (beta)/( x^(2))]` should also be `M^(1) L^(2)T^(-3)` `[beta] = M^(1) L^(4) T^(-3) and [Fv + ( beta)/( x^(2))]` will also have dimension `M^(1)L^(2) T^(-3)`. So, `([alpha)]/([ t^(2)]) = M^(1) L^(2)T^(-3) , [alpha] = M^(1) L^(2) T^(-1)` |
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| 38. |
The energy `E` of an oscillating body in simple harmonic motion depends on its mass `m`, frequency `n` and amplitude `a` using the method of dimensional analysis find the relation between E ,m, n and a .A. `M f//A^(2)`B. `M fA^(-2)`C. ` Mf^(2) A^(-2)`D. ` MF^(2) A^(2)` |
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Answer» Correct Answer - D `E prop m^(a) f^(b) A^(c )` `ML^(2) T^(-2) = m^(a) T^(-b) L^( c )` `:. a = 1 , b = 2 , c = 2` |
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| 39. |
The frequency `f` of vibrations of a mass `m` suspended from a spring of spring constant `k` is given by `f = Cm^(x) k^(y)` , where `C` is a dimensionnless constant. The values of `x and y` are, respectively,A. `(1)/(2) , (1)/(2)`B. ` -(1)/(2) , -(1)/(2)`C. `(1)/(2) , - (1)/(2)`D. `- (1)/(2) , (1)/(2)` |
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Answer» Correct Answer - D `f = Cm^(x)k^(y)`. Writing dimensions on both sides. `[M^(0)L^(0)T^(-1)] = M^(x) [ML^(0)T^(-2)]^(y) = [M^(x+y) T^(-2 y)]` Comparing dimensions on both sides , we have ` 0 = x + y and -1 = -2 y rArr y = (1)/(2) , x = -(1)/(2)` Aliter. Remembering that the frequency of oscillation of loaded spring is ` f = (1)/( 2 pi) sqrt(k)/( m) = (1)/( 2 pi) (k)^(1//2) m^(-1//2)` which gives `x = -(1)/(2) and y = (1)/(2)` |
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| 40. |
A body of mass `m` hung at one end of the spring executes simple harmonic motion . The force constant of a spring is `k` while its period of vibration is `T`. Prove by dimensional method that the equation `T = 2 pi m // k` is correct. Dervive the correct equation , assuming that they are related by a power law. |
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Answer» Correct Answer - ` C sqrt((m)/(k))` The given equation is `T = ( 2 pi m)/( k)` Taking the dimensions of both sides, we have `[T] = ([M])/( [ ML^(0)T^(-2)] = T^(2)` As the dimensions of two sides are not equal , hence the equation is incorrect. Let the correct relation be `T = Cm^(a) k^(b) , where C` is constant. Equating the dimensions of both sides , we get `[T] = [M]^(a) [MT^(-2)]^(b)` or `[M^(0) L^(0) T] = [M^(a+b)L^(0)T^(-2b)]` Comparing the powers of M,L, and T on both sides , we get ` a + b = 0 and -2b = 1`. Therefore , `b = -(1)/( 2) and a = (1)/(2)` :. `T = C m^(-1//2) k ^(-1//2) = C sqrt((m)/(k))` This is the correct equation. |
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| 41. |
The momentum of inertia of a body rotating about a given axis is `12.0 kg m^(2)` in the SI system . What is the value of the moment of inertia in a system of units in which the unit of lengths is `5 cm` and the unit of mass is `10 g`?A. ` 2.4 xx 10^(3)`B. `6.0 xx 10^(3)`C. `5.4 xx 10^(5)`D. `4.8 xx 10^(5)` |
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Answer» Correct Answer - D ` n_(2) = n_(1) ((M_(1))/(M_(2)))^(a) ((L_(1))/(L_(2)))^(b) ((T_(1))/(T_(2)))^( c )` Dimensional formula of moment of interia ` = [ML^(2) T^(0)]` :. ` a = 1, b= 2 , c = 0 ` `n_(1) = 12.0 , M_(1) = 1 kg , M_(2) = 10 g ` `L_(1) = 1 m , L_(2) = 5 cm , T_(1) = 1 s, T_(2) = 1 s` `n_(2) = 12.0 ((1 kg)/( 10 g)) ^(1) ((1 m)/( 5 cm))^(2) ((1 s)/( 1s))^(0)` ` = 12 xx ((1000 g )/( 10 g)) ^(1) (( 100 cm) /(5 cm))^(2) xx 1` ` = 12 xx 100 xx 400 = 4.8 xx 10^(5)` |
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| 42. |
The number of particles is given by `n = -D(n_(2) - n_(1))/( x_(2) - x_(1))` crossing a unit area perpendicular to X - axis in unit time , where `n_(1)and n_(2)` are particles per unit volume for the value of `x` meant to `x_(2) and x_(1)` . Find the dimensions of `D` called diffusion constant. |
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Answer» (n) = Number of particle passing from unit area in unit time `= ("Number of particles")/( Axxt) = ([ M^(0) L^(0)T^(0)])/([L^(2)] [ T]) = [L^(-2)T^(-1)]` `[n_(1)] = [ n_(2)]` = Number of particles in unit volume = `[L^(-3)]` Now from the given formula , `[D] = ([n] [ x_(2) - x_(1)])/([ n_(2) - n_(1)]) = ([L^(-2)T^(-1)][L])/([L^(-3)]) = [L^(2)T^(-1)]` |
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| 43. |
The number of particles is given by `n = -D(n_(2) - n_(1))/( x_(2) - x_(1))` crossing a unit area perpendicular to X - axis in unit time , where `n_(1)and n_(2)` are particles per unit volume for the value of `x` meant to `x_(2) and x_(1)` . Find the dimensions of `D` called diffusion constant.A. `[M^(0) L T^(-2)]`B. `[M^(0) L^(2) T^(-4)]`C. `[M^(0) L^(2) T^(-2)]`D. `[M^(0) L^(2) T^(-1)]` |
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Answer» Correct Answer - D ` n = -(D(n_(2)-n_(1)))/(x_(2)-x_(1)) rArr T^(-1) L^(-2) = ( D(L^(-3)))/(L) rArr D = (T^(-1)L^(-2) xx L)/(L^(-3))` `rArr D = [ M^(0) L^(2) T^(-1)]` |
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| 44. |
(a). Calculate the area enclosed by a circle of radius `0.56 m` to the correct number of significant figures. (b) . Calculate the area enclosed by a circle of diameter `1.12 m` to the correct number of significant figures. |
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Answer» Correct Answer - `~ 0.99 m^(2)` `A = pi r^(2) = (22)/( 7) ( 0.56)^(2) = 0.9856 m^(2) ~ 0.99 m^(2)` (two significant figures) |
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| 45. |
The number of significiant figures in `5.69 xx 10^(15) kg` isA. `1`B. `2`C. `3`D. `18` |
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Answer» Correct Answer - C ` 5.69 xx 10^(15) kg` has three significant figures as the power of `10 ` is not considered for significant figures. |
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| 46. |
The period of oscillation of a simple pendulum is `T = 2 pi sqrt((L)/(g)) .L` is about `10 cm` and is known to `1mm` accuracy . The period of oscillation is about `0.5 s`. The time of 100 oscillation is measured with a wrist watch of `1 s` resolution . What is the accuracy in the determination of `g` ? |
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Answer» Correct Answer - `5%` `T = 2 pi sqrt((L)/(g))` or `T^(2) = 4 pi^(2) (L)/( g)` or `g = ( 4 pi^(2) L)/( T^(2))` Now `(Delta g)/( g) = (Delta L) /( L) + 2 xx(Delta T)/(T)` IN terms of percentage , `(Delta g)/(g) xx 100 = (Delta L)/(L) xx 100 + 2 xx (Delta T)/(T) xx 100` Percentage error in `L = 100 xx ( Delta L)/(L) = 100 xx (0.1)/(10) = 1%` Percentage error in `T = 100 xx ( Delta T%)/(T) = 100 xx (1)/(50 ) = 2%` Percentage error in `g = 100 xx ( Delta g)/(g) = 1% + 2 xx 2% = 5%` |
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| 47. |
The order of magnitude of `499 is 2`, then the order of magnitude of `501` will beA. `1`B. `2`C. `1`D. `3` |
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Answer» Correct Answer - D `501 = 0.501 xx 10^(3) rArr` order of magnitudde of 501 is 3. |
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| 48. |
(a). Two plates have lengths measured as `( 1.9 +- 0.3) m and ( 3.5 +- 0.2)m`. Calculate their combined length with error limits. (b) The initial and final temperatures of a liquid are measured to be `67.7 +- 0.2^(@) C and 76.3 +- 0.3^(@) C`. Calculate the rise in temperature with error limits. |
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Answer» Correct Answer - (a). ` ( 5.4 +- 0.5 ) m`; (b). ` = 8.6 +- 0.5^(@) C` (a). `l = l_(1) + l_(2) = ( 1.9 +- 0.3) + (3.5 +- 0.2) = (5.4 +- 0.5) m` (b). `Delta T = T_(2) - T_(1) = ( 76.3 +- 0.3) - ( 67.7 +- 0.2)` ` = 8.6 +- 0.5^(@) C` |
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| 49. |
The pressure on a square plate is measured by measuring the force on the plate and the length of the sides of the plate . If the maximum error in the measurement of force and length are , respectively , `4% and 2%`. Find the maximum error in the measurement of pressure. |
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Answer» `P = (F)/(A) = (F)/(l^(2))`, so maximum error in pressure `(P)` `((Delta P)/(P) xx 100)_(max) = (Delta F)/(F) xx 100 + 2 (Delta l)/(l) xx 100` ` = 4% + 2 xx 2% = 8%` |
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| 50. |
The resistance `R = V//i`, where `V = 100 +- 5 V and I = 10+- 0.2 A`. What is the total error in `R`? |
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Answer» `R = (V)/(I) (( Delta R )/( R ) xx 100)_(max) = (Delta V)/(V) xx 100 + (Delta I)/(I) xx 100 ` ` = (5)/(100) xx 100 + (0.2)/(10) xx 100` `= ( 5+2)% = 7%` |
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