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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.
| 301. |
The percentage error in the measurement of the radius of a sphere is 1.5%. What would be the percentage error in the volume of the sphere ? |
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Answer» Let the radius of the sphere=`r pm Deltar ` Then according to the question , `(Deltar)/rxx100`=1.5% or`(Deltar)/r`=0.015 Volume of the sphere is `V=4/3pir^3` As per the rule for combination of errors for exponents `(DeltaV)/V=3(Deltar)/r` =3 x 0.015 `(DeltaV)/V`=0.045 Percentage error =`(DeltaV)/Vxx100=0.045xx100=4.5%` |
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| 302. |
If the error in the measurement of radius of a sphere in `2%` then the error in the determination of volume of the spahere will beA. 0.02B. 0.04C. 0.06D. 0.08 |
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Answer» Correct Answer - C |
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| 303. |
A physical quantity y is given by `y=(P^2Q^(3//2))/(R^4S^(1//2))` The percentage error in A,B , C and D are 1%,2%, 4% and 2% respectively. Find the percentage error in y. |
| Answer» Correct Answer - `22%` | |
| 304. |
A vernier callipers is used to measure the width and the length of a rectangular plate. The measured values are 1.38 cm and 4.02 cm respectively.Find the uncertainly in the value of its area. |
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Answer» Given that, width w=1.38 cm length , l=4.02 cm Since the least count of a vernier callipers is 0.01 cm, each of the above mentioned measurements has an uncertainly of `pm`0.01 cm. hence, the values can be written as w=1.38 cm `pm`0.01 cm `l=4.02 cm pm 0.01 cm` The area obtained using the measured value is A=w x l = 1.38 x 4.02 `A=5.55 cm^2` ....(i) If we take the lower limits of both the measured values i.e. `l_"min"`=4.02-0.01 =4.01 cm and `w_"min"`=1.38-0.01 =1.37 cm The minimum possible area (lower limits of the area ) comes out to be `A_"min"=4.01xx1.37 cm^2` `A_"min"=5.49 cm^2`...(ii) The upper limit can also be calculated similarly. `therefore A_"max"=4.03xx1.39 cm^2` `A_"max"=5.60 cm ^2` ...(iii) using equation (i),(ii) and (iii) , we find that `A_"min"` is lower than A (the nominal measurement ) by the value `0.06 cm^2` And `A_"max"` is higher than A by the value `0.05 cm^2` As a practical rule, we choose the higher of these two deviations (from the measured value ) as the uncertainly, in our result. Therefore, in this particular example, we can write the area of plate as `5.55 cm^2 pm 0.06 cm^2` |
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| 305. |
How many significant figures are there in the measured values (i)227.2 g , (ii)3600 g and (iii)0.00602 g |
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Answer» (i)227.2 g has all the non-zero digits. Hence, it has four significant figures. (ii)According to rule number 3, trailing zeroes are not significant. hence, 3600 g has 2 significant figures. (iii)0.00602 g According to the rule number 5, the zeroes at the beginning are not significant . Hence, 3 significant figures. |
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| 306. |
`SI` Unit of physical quantity whose dimensional formula is `M^(-1) L^(-2) T^(4) A^(2)` isA. ohmB. voltC. siemenD. farad |
| Answer» Correct Answer - D | |
| 307. |
If energy `E`, velocity `v` and time `T` are taken as fundamental quanties, the dimensional formula for surface tension isA. `[EV^(-2)T^(-2)]`B. `[E^(-2)VT^(-2)]`C. `[E^(-2)V^(-2)T]`D. `[E^(-2)V^(-2)T^(-2)]` |
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Answer» Correct Answer - A |
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| 308. |
For the equation `F = A^(a) v^(b) d^(c)` where `F` is force, `A` is area, `v` si velocity and `d` is density with the dimensional anaysis gives the following values for the exponents.A. `a = 1, b = 2, c = 1`B. `a = 2, b = 1, c = 1`C. `a = 1, b = 1, c = 2`D. `a = 0,b = 1, c = 1` |
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Answer» Correct Answer - A `F = A^(a) v^(b) d^(c), MLT^(-2) = (L^(2)) (LT^(-1))^(b) (ML^(-3))^(c)` comparing the powers on both sides |
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| 309. |
The radius of a sphere is `(5.3 +- 0.1)`cm` The perecentage error in its volume isA. `0.1/5.3xx100`B. `3xx0.1/5.3xx100`C. `3/2xx0.1/5.3xx100`D. `6xx0.1/0.3xx100` |
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Answer» Correct Answer - B |
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| 310. |
A public park , in the form of a square , has an area of `(100pm0.2 ) m^2` The side of park isA. (10 `pm` 0.01) mB. (10 `pm` 0.1) mC. (10 `pm` 0.02) mD. (10 `pm` 0.2) m |
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Answer» Correct Answer - A |
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| 311. |
Frequency of sound that can be produced by a pipe depends on length (l) of the pipe, atmospheric pressure (p) and density (d) of air, according to relation `v=(p^b d^c)/t^a` . Find the value of (a +b + c). |
| Answer» Correct Answer - 1 | |
| 312. |
In a new system, the unit of mass is made 10 times, the unit of length is made 1/100 times and unit of time is made 10 times. Magnitude of 1J in the new system of unit is `10^x`.What is the value of x ? |
| Answer» Correct Answer - 5 | |
| 313. |
`[M^(1) L^(2) T^(-3) A^(-2)]` si the dimensional formula of:A. electric resistanceB. capacityC. electric potentialD. specific resistance |
| Answer» Correct Answer - A | |
| 314. |
Find the maximum possible percentage error in the measurement of force on an object (of mass m) travelling at velocity v in a circle of radius r, if m= (4.0 `pm` 0.1)kg, v = (10 `pm` 0.1) m/s and r=(8.0`pm`0.2)m. |
| Answer» Correct Answer - 7 | |
| 315. |
A : Higher is the accuracy of measurement, if instrument have smaller least count. R : Smaller the percentage error, higher is the accuracy of measurement.A. If both Assertion & Reason are true and the reason is the correct explanation of the assertion, then mark (1).B. If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark (2).C. If Assertion is true statement but Reason is false, then mark (3).D. If both Assertion and Reason are false statements, then mark (4). |
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Answer» Correct Answer - B |
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| 316. |
`P = (nx^(y)T)/(V_(0))e^(-(Mgh)/(nxT))`, where n is number of moles, P is represents acceleration due to gravity and h is height. Find dimension of x and value of y. |
| Answer» Correct Answer - `[ML^2T^(-2)mu^(-1)K^(-1)] [ M^0 L^(-0)T^0]` | |
| 317. |
The percentage error in the measurement of mass and speed of a body are 2% and 3% respectively. What will bek the maximum percentage error in the estimation of kinetic energy of the body?A. 0.02B. 0.01C. 0.05D. 0.07 |
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Answer» Correct Answer - D |
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| 318. |
The percentage error in the measurement of the voltage V is 3% and in the measurement of the current is 2%. The percentage error in the measurement of the resistance isA. 0.03B. 0.02C. 0.01D. 0.05 |
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Answer» Correct Answer - D |
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| 319. |
What are the units of `K (1)/(4pi in_(0))` ?A. `C^(2) N^(-1) m^(-2)`B. `C^(-2) N^(-1) m^(2)`C. `C^(2) N^(1) m^(2)`D. unitless |
| Answer» Correct Answer - B | |
| 320. |
As student performs an experiment for determine of `g[=(4pi^(2)L)/(T^(2))]. L approx 1m`, and has commits an error of `Delta L` for `T` he tajes the teime of `n` osciollations wityh the stop watch of least count `Delta T`. For which of the following data the measurement of `g` will be most accurate?A. `Delta L = 0.5, DeltaT = 0.1n = 20`B. `Delta L = 0.5, DeltaT = 0.1n = 50`C. `Delta L = 0.5, DeltaT = 0.01n = 20`D. `Delta L = 0.5, DeltaT = 0.05n = 50` |
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Answer» Correct Answer - D `(Delta g)/(g) = (Delta l)/(l) + 2(Delta T)/(T)` (`Delta l` and `DeltaT` are least and the number of readings are maximum) |
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| 321. |
Speed of light in SI system is `3xx10^8` m/s. What is the speed of light in a new system of units in which unit of length is x km and unit of time is Y millisecond? |
| Answer» Correct Answer - `3xx10^2 Y/x`units | |
| 322. |
What are the unit of `(mu_(0))/(4pi)`A. `NA^(-1) m^(2)`B. `NA^(-2)`C. `Nm^(2) C^(2)`D. unitless |
| Answer» Correct Answer - B | |
| 323. |
Find the maximum possible percentage error in the measurement of force on an object (of mass m) travelling at velocity v in a circle of radius r if m =`(4.0 pm 0.1)`kg, v= `(10pm0.1)` m/s and r=`(8.0 pm 0.2)` m |
| Answer» Correct Answer - 7 | |
| 324. |
Spring constant of a spring is calculated using formule `K=(4pi^2M)/T^2`, where T is time period of vertical oscillation when mass M is hung with the help of spring to rigid support. If time of oscillation for 10 oscillations is measured to be 5.0 s and mass M=0.20 kg, find possible error in spring constant K. |
| Answer» Correct Answer - 2.9 N/m | |
| 325. |
The velocity of water wave `v` may depend on their wavelength `lambda`, the density of water `rho` and the acceleration due to gravity `g`. The method of dimensions gives the relation between these quantities as |
| Answer» Correct Answer - `v=ksqrt(lambdag)` | |
| 326. |
The time period Of oscillation of a simple pendulum depends on the following quantities Length of the pendulum (l), Mass of the bob (m), and Acceleration due to gravity (g) Derive an expression for Using dimensional method. |
| Answer» Dimensionless constant =`2pi` or `T=2pisqrt(l/g)` | |
| 327. |
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 of |
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Answer» D.F of `(EJ^(2))/(M^(5) G^(2))` Substituing `D.F` of `E,J,M` and `G` in above formula `= (ML^(2)T^(-2)[ML^(2)T^(-1)]^(2))/(M^(5)[M^(-1)L^(3)T^(-2)]^(2)) = [M^(0) L^(0) T^(0)]` |
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| 328. |
A screw gauge gives the following reading when used to mesure the diametre of a wire. Main scale reading : `0mm` Circular scale reading : `52 divisions` Given that `1mm` on main scale corresponds to `100` divisions of the circular scale. the diameter of wire from the above data is : |
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Answer» Main scale reading `= 0mm` Circular scale reading `= 52` divisions Least count `= ("Value of 1 main scale division")/("Total divisions on circular scale") = (1)/(100) mm` Diameter of wire `= M.S.R + (C.S.R xx L.C)` `= 0 + 52 xx (1)/(100) mm = 0.52mm = 0.52cm` |
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| 329. |
The diameter of a wire as measured by a screw gauge was found to be `1.002cm, 1.004cm` and `1.006cm`. The absoulue error in the third reading isA. `0.002cm`B. `0.004cm`C. `1.002cm`D. Zero |
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Answer» Correct Answer - A `Delta x_(3) = |x_(3) - x_("mean")|` |
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| 330. |
The least count of a stop watch is `(1//5)s`. The time 20 oscillations of a pendulum is measured to be `25 s`. The maximum percentage error in this measurement isA. `0.1%`B. `0.8%`C. `1.8%`D. 0.08 |
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Answer» Correct Answer - B |
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| 331. |
Which of the following time measuring devices is most precise ? (a) A wall clock. (b) A stop watch. (c ) A digital watch. (d) An atomic clock. Give reason for you answer. |
| Answer» A wall clock can measure time correctly upto one second. A stop watch can measure time correctly upto a fraction of a second. A digital watch can measure time up to a fraction of second. An atomic clock can measure time most precisely as its precision is 1 s in `10^(13)` s. | |
| 332. |
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 figures it would be `2.74` Rounding off `2.735` to `3` significant figures it would be `2.74` |
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| 333. |
The time dependence of a physical quantity `P` is given by `P = P_(0)e^(-alpha t^(2))` , where `alpha` is a constant and `t` is time . Then constant `alpha` is//hasA. [T]B. `[T^(2)]`C. `[T^(-1)]`D. `[T^(-2)]` |
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Answer» Correct Answer - D |
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| 334. |
The diamensions of time in energy are |
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Answer» Correct Answer - B |
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| 335. |
The time period of oscillation of a body is given by `T=2pisqrt((mgA)/(K))` K: Represents the kinetic energy, m mass, g acceleration due to gravity and A is unknown If `[A]=M^(x)L^(y)T^(z)`, then what is the value of x+y+z?A. 3B. 2C. 1D. 5 |
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Answer» Correct Answer - A `[T]=[((mgA)/(K))^(1//2)]=[(mgA)/((mv^(2)))]^(1//2)=[(A)/(vT)]^(1//2)=[(A)/(L)]^(1//2)` `implies[A]=[M^(0)LT^(2)]implies[M^(x)L^(y)T^(z)]=[M^(0)LT^(2)]` `impliesx=0,y=1,z=2impliesx+y+z=3` |
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| 336. |
Assertion : Absolute error may be negative or positive. Reason : Absolute error is the difference between the real value and the measured value of a physical quantity.A. If both Assertion & Reason are true and the reason is the correct explanation of the assertion, then mark (1).B. If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark (2).C. If Assertion is true statement but Reason is false, then mark (3).D. If both Assertion and Reason are false statements, then mark (4). |
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Answer» Correct Answer - D |
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| 337. |
The mass and volume of a body are 4.237 g and `2.5cm^3,` respectively. The density of the meterial of the body in correct significant figures isA. `1.6948gcm^(-3)`B. `1.69gcm^(-3)`C. `1.7gcm^(-3)`D. `1.695gcm^(-3)` |
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Answer» Correct Answer - C Here, mass, `m=4.237g` Volume, `V=2.5cm^(3)` Density `rho=("mass")/("Volume")=(4.237g)/(2.5cm^(3))=1.6948cm^(-3)` As mass has 4 significant figures and volume has 2 significant figures, therefore, as per rule, density will have only two significant figures. Rounding off to two significant figures, we get `rho=1.7g" "cm^(-3)` |
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| 338. |
The mass and volume of a body are 4.237 g and `2.5cm^(3)` respectively. The density of the material of the body in correct significant figures isA. `1.6048g cm^(-3)`B. `1.69 cm^(-3)`C. `1.7g cm^(-3)`D. `1.695g cm^(-3)` |
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Answer» Correct Answer - C In this question, density should be reported to two signigcant figures Density `= (4.237g)/(2.5cm^(3)) = 1.6948` As rounding off the numbers, we get density `= 1.7` |
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| 339. |
If length and breath of a plate are `(40 +- 0.2)cm` and `(30 +- 0.1 ) cm`, the absolute error in the meaurement of area isA. `10 cm^(2)`B. `8 cm^(2)`C. `9cm^(2)`D. `7cm^(2)` |
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Answer» Correct Answer - A `A = lb implies (Delta A)/(A) = (Delta l)/(l) + (Delta b)/(b) implies Delta A = A [(Delta l)/(l) + (Delta b)/(b)]` `Delta A = b Delta l + l Delta b = 10cm^(2)` |
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| 340. |
The length and breath of a recantangular object are `25.2cm` and `16.8cm` respecitively and have been measured to an accuracy of `0.1cm`. Relative error and percentage error in the area of the object areA. `0.1 & 1%`B. `0.02 & 2%`C. `0.03 & 3%`D. `0.4 & 4%` |
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Answer» Correct Answer - A `A = l xx b , (Delta A)/(A) = (Delta l)/(l) + (Delta b)/(b)` `(Delta A)/(A) xx 100 = ((Delta l)/(l) + (Delta b)/(b)) xx 100` |
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| 341. |
A recantanglar metal slab of mass `33.333` has its length `8.0cm`, breath `5.0cm` and thickness `1mm`. The mass is measured with accuracy up to `1mg` with a senstive balance. The length and breath are measured with vernier calipers having a least count of `0.01cm`. The thickness is measured with a new a screw gauge of least count `0.01mm`. The percentage accuracy in density calculated from the above measurements isA. `13%`B. `130%`C. `1.6%`D. `16%` |
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Answer» Correct Answer - C Percentage error gives percentage accuracy `d = (m)/(lbh)` relative error, `(Delta d)/(d) = (Delta m)/(m) + (Delta l)/(l) + (Delta b)/(b) + (Delta h)/(h)` and calculate `((Delta d)/(d)) xx100` |
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| 342. |
The length and breadth of a rectangular plate are measured to be 14.5 cm and 4.2cm respectively. Find its area to appropriate significant figures. |
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Answer» Here length = 14.5 cm `to` 3 significant figures. breadth =4.2 cm `to` 2 significant figures. `therefore` Area = `14.5xx4.2 cm^2` `=60.90 cm^2` The answer should be rounded off to two significant figures. `therefore` The digit to be dropped in the answer 60. 9 0 is 9 (circled ) which is greater than 5. So the preceding zero is raised by 1. hence the answer `61 cm^2` |
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| 343. |
An object of mass 4.237 g occupies a volume `1.72 cm^3`. The density of the object to appropriate significant figures is -A. `2.46 "g cm"^(-3)`B. `2.463 "g cm"^(-3)`C. `2.5 "g cm"^(-3)`D. `2.50 "g cm"^(-3)` |
| Answer» Correct Answer - 1 | |
| 344. |
The length, breath and thicknes of a recantagular lamina are `1.024m, 0.56m`, and `0.0031m`. The volume is .......`m^(3)`A. `1.8xx10^(-3)`B. `1.80xx10^(-3)`C. `0.180xx10^(-4)`D. `0.00177` |
| Answer» Correct Answer - A | |
| 345. |
Assertion: A number 2.746 rounded off to three significant figures is 2.75, while the number 2.743 would be 2.74. Reason: In rounding off the uncertain digits, the preceding digit is raised by 1 if the insignificant digit to be dropped is more than 5 and is left unchanged if the latter is less than 5.A. if both assertion and reason are true reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation fo assertion.C. If assertion is true but reaso is false.D. IF both assertion and reason are false. |
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Answer» Correct Answer - A a |
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| 346. |
Round off the following numbers upto three significant figures. (i)2.520 , (ii)4.645 , (iii)22.78 , (iv)36.25 |
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Answer» (i)2.520 Since 0 is less than 5, prededing digit is left unchanged . Hence , 2.52 (ii)4.645 since the digit to be dropped is 5 and preceding digit 4 is even .Hence , 4.64 (iii)22.78 Since the digit to be dropped, 8 is greater than 5 , the preceding digit 7, is raised by 1. Hence 22.8 (iv)36.35 Since the digit to be dropped is 5, and the preceding digit 3 is odd, we write the answer as 36.4 |
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| 347. |
STATEMENT-1 The mass of an object is 132 kg. The number of significant figures in this measurement is 3 STATEMENT 2 The same mass when expressed in grams as 132000 g has six significant figuresA. Statement-1 is True , Statement-2 is True ,Statement-2 is a correct explanation for Statement-1B. Statement-1 is True , Statement-2 is True ,Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True , Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - 3 | |
| 348. |
When a force is expressed in dyne, the number of sinificant figures is four. If it is expressed in newton, the number of significant figures will become `(10^(5) "dyne" = 1N)`A. `9`B. `5`C. `1`D. `4` |
| Answer» Correct Answer - D | |
| 349. |
The velocity of light in vacumm is 30 crore `m//s`. This is expressed is standard form up to 3 significant figures asA. `0.03xx10^(11) m//s`B. `300xx10^(6) m//s`C. `3.00xx10^(8) m//s`D. `0.030xx10^(10) m//s` |
| Answer» Correct Answer - C | |
| 350. |
The numbers 3.845 and 3.835 on rounding off to 3 significant figures will giveA. 3.85 and 3.84B. 3.84 and 3.83C. 3.85 and 3.83D. 3.84 and 3.84 |
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Answer» Correct Answer - D The number 3.845 rounded off to three significant figrues becomes 3.84 since the preceeding digit is even, on the other hand, the number 3.835 rounded off to three significant figures becomes 3.84 since the preceeding digit is odd. |
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