Q 1 :

In an experiment to measure focal length (f) of convex lens, the least count of the measuring scales for the position of object (u) and for the position of image (v) are u and v, respectively. The error in the measurement of the focal length of the convex lens will be                            [2024]

  • 2f[uu+vv]

     

  • f[uu+vv]

     

  • uu+vv

     

  • f2[uu2+vv2]

     

(4)

       1v-1u=1f

       -1v2dv+1u2du=-1f2df dff2=dvv2+duu2

       df=f2[dvv2+duu2]f=f2(vv2+uu2)

 



Q 2 :

To find the spring constant (k) of a spring experimentally, a student commits 2% positive error in the measurement of time and 1% negative error in measurement of mass. The percentage error in determining value of k is:                                    [2024]

  • 3%

     

  • 4%

     

  • 5%

     

  • 1%

     

(1)  

        T=2πmkk=4π2·mT2

        kk=mm+2·TTkk%=mm%+2TT%

        (We always take maximum error)

        kk%=(-1)%+2(2)%=|-3%|=3%



Q 3 :

Young's modulus is determined by the equation given by Y=49000mLdynecm2 where M is the mass and L is the extension of wire used in the experiment. Now error in Young's modules (Y) is estimated by taking data from M-L plot in graph paper. The smallest scale divisions are 5g and 0.02 cm along load axis and extension axis respectively. If the value of M and are 500 g and 2 cm respectively then percentage error of Y is                      [2024]

  • 0.2%

     

  • 0.02%

     

  • 2%

     

  • 0.5%

     

(3) 

        YY=mm+ll

        =5500+0.022=0.01+0.01YY

        %YY=2%



Q 4 :

The resistance R=V/I where V=(200±5)V and I=(20±0.2)A. The percentage error in the measurement of R is:              [2024]

  • 3.5%

     

  • 7%

     

  • 3%

     

  • 5.5%

     

(1) 

      We have, R=V/I

       According to error analysis, maximum fractional error.

       dRR=dVV+dII

       dRR=5200+0.220=7200

        Hence, % error dRR×100=7200×100=3.5%



Q 5 :

A physical quantity Q is found to depend on quantities a,b,c by the relation Q=a4b3c2. The percentage error in a,b and c are 3%, 4% and 5% respectively. Then, the percentage error in Q is                                       [2024]

  • 66%

     

  • 14%

     

  • 34%

     

  • 43%

     

(3)  

      QQ=4aa+3bb+2cc

      Q=4×3%+3×4%+2×5%

       =12%+12%+10%=34%

        

 



Q 6 :

If the percentage errors in measuring the length and the diameter of a wire are 0.1% each. The percentage error in measuring its resistance will be          [2024]

  • 0.2%

     

  • 0.144%

     

  • 0.3%

     

  • 0.1%

     

(3)  

        R=ρLπd24

        RR=LL+2dd

        LL=0.1% and dd=0.1%

         RR=0.3%



Q 7 :

The measured value of the length of a simple pendulum is 20 cm with 2 mm accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is N%. The value of N is:                    [2024]

  • 4

     

  • 8

     

  • 6

     

  • 5

     

(3)  

       T=2πlg

        g=4π2lT2

        gg=ll+2TT=0.220+2(140)=1.220

        Percentage change = 1.220×100=6%



Q 8 :

The radius (r), length (l) and resistance (R) of a metal wire was measured in the laboratory as                              [2024]

 

r=(0.35±0.05)cm

 

R=(100±10)ohm

 

l=(15±0.2)cm

 

The percentage error in resistivity of the material of the wire is

  • 25.6%

     

  • 39.9%

     

  • 37.3%

     

  • 35.6%

     

(2) 

        ρρ=RR+2rr+ll

        =10100+2×0.050.35+0.215

         =110+27+175

          ρρ=39.9%



Q 9 :

The maximum percentage error in the measurement of density of a wire is

[Given, mass of wire = (0.60 ± 0.003) g

radius of wire = (0.50 ± 0.01) cm

length of wire = (10.00 ± 0.05) cm]           [2025]

  • 4

     

  • 5

     

  • 8

     

  • 7

     

(2)

ρ=mvol.=mπR2l  dρρ=dmm+2dRR+dll

 dρρ=(0.0030.6+2×0.010.5+0.0510)100=5%



Q 10 :

The energy of a system is given as E(t)=α3eβt, where t is the time and β=0.3 s1. The error in the measurement of α and t are 1.2% and 1.6%, respectively. At t = 5 s, maximum percentage error in the energy is          [2025]

  • 4%

     

  • 11.6%

     

  • 6%

     

  • 8.4%

     

(3)

dE=3α2dαeβt+α3eβt(β)dt

dEE=3dαα+(β)dt

EE=3αα+βt

(%)T=1.6×5=8%

%(3αα)=3×1.2%=3.6%

Putting values we get,

EE%=3.6+8×0.3=6%.



Q 11 :

A quantity Q is formulated as X2Y+32Z25·X,Y, and Z are independent parameters which have fractional errors of 0.1, 0.2 and 0.5, respectively in measurement. The maximum fractional error of Q is          [2025]

  • 0.1

     

  • 0.8

     

  • 0.7

     

  • 0.6

     

(3)

Fractional error =2XX+32YY+25ZZ

                                      =2(0.1)+32(0.2)+25(0.5)

                                       = 0.2 + 0.3 + 0.2 = 0.7



Q 12 :

A physical quantity Q is related to four observables a, b, c, d as follows:

Q=ab4cd where, a=(60±3) Pa; b=(20±0.1) m,

c=(40±0.2) Nsm2 and d=(50±0.1) m,

then the percentage error in Q is x1000, where x = __________.          [2025]



(77)

Q=ab4cd

 QQ=[aa+4bb+cc+dd]

 x1000=[360+4(0.120)+(0.240)+0.150]

 x=77



Q 13 :

A physical quantity C is related to four other quantities p, q, r and s as follows

               C=pq2r3s

The percentage errors in the measurement of p, q, r and s are 1%, 2%, 3% and 2% respectively. The percentage error in the measurement of C will be __________ %.          [2025]



(15)

C=p1q2r3s1/2

(dCC)max=dPP+2dqq+3drr+12dss

                       =(1+2×2+3×3+12×2)%=15%



Q 14 :

Two resistances are given as R1=(10±0.5)Ω and R2=(15±0.5)Ω. The percentage error in the measurement of equivalent resistance when they are connected in parallel is        [2023]

  • 6.33

     

  • 5.33

     

  • 2.33

     

  • 4.33

     

(4)

1R=1R1+1R2

Differentiating both sides, we get

RR2=R1R12+R2R22[R=R1R2R1+R2=10×1510+15=6]

 RR=(R1R12+R2R22)R

 (0.5100+0.5225)6=(6×0.525)(14+19)=13100

RR×100=133=4.33%



Q 15 :

A 2 meter long scale with least count of 0.2 cm is used to measure the locations of objects on an optical bench. While measuring the focal length of a convex lens, the object pin and the convex lens are placed at 80 cm mark and 1 m mark, respectively. The image of the object pin on the other side of lens coincides with image pin that is kept at 180 cm mark. The % error in the estimation of focal length is          [2023]

  • 1.02

     

  • 0.85

     

  • 1.70

     

  • 0.51

     

(3)

Least count of the scale is 0.2 cm

u = (100 ± 0.2) – (80 ± 0.2) = (20 ± 0.4) cm

v = (180 ± 0.2) – (100 ± 0.2) = (80 ± 0.4) cm

From lens formula,

1v1u=1f  1f=180120  f=16 cm

Also, vv2+uu2=ff2

 ff×100=(vv2+uu2)×f×100

 %f=(0.4400+0.46400)×16×100=1.70

 



Q 16 :

A cylindrical wire of mass (0.4 ± 0.01) g has length (8 ± 0.04) cm and radius (6 ± 0.03) cm. The maximum error in its density will be         [2023]

  • 1%

     

  • 4%

     

  • 5%

     

  • 3.5%

     

(2)

ρ=mπr2l  |dρρ|max=|dmm|+2|drr|+|dll|

                                                =0.010.4+2(0.03)6+0.048

 % error in density = (dρρ)×100%

                                               =(2.5+1+0.5)%=4%



Q 17 :

A physical quantity P is given as P=a2b3cd. The percentage error in the measurement of a, b, c and d are 1%, 2%, 3% and 4% respectively. The percentage error in the measurement of quantity P will be          [2023]

  • 13%

     

  • 14%

     

  • 12%

     

  • 16%

     

(1)

PP×100%=(2aa+3bb+cc+12dd)×100%

                                =2(1%)+3(2%)+3%+12×4%=13%



Q 18 :

A body of mass (5 ± 0.5) kg is moving with a velocity of (20 ± 0.4) m/s. Its kinetic energy will be          [2023]

  • (1000 ± 140) J

     

  • (1000 ± 0.14) J

     

  • (500 ± 0.14) J

     

  • (500 ± 140) J

     

(1)

k=12mv2

k=12×5×400=5×200=1000 J

kk=mm+2vv=0.55+2×0.420

k=1000(110+4100)=1000(10+4100)=140 J



Q 19 :

Four persons measure the length of a rod as 20.00 cm, 19.75 cm, 17.01 cm, and 18.25 cm. The relative error in the measurement of average length of the rod is:      [2026]

  • 0.08

     

  • 0.18

     

  • 0.24

     

  • 0.06

     

(4)

mean=1+2+3+44

mean=20.00+19.75+17.01+18.254

= 18.75

Δmean=|Δ1|+|Δ2|+|Δ3|+|Δ4|4

=1.25+1+1.74+0.54=1.12

So, relative error

=Δmeanmean=1.1218.75= 0.06



Q 20 :

In an experiment the values of two spring constants were measured as k1=(10±0.2) N/m and k2=(20±0.3) N/m. If these springs are connected in parallel, then the percentage error in the equivalent spring constant is:   [2026]

  • 1.67%

     

  • 2.67%

     

  • 2.33%

     

  • 1.33%

     

(1)

For parallel combination of spring,

Keq=K1+K2=30 N/m

ΔKeq=ΔK1+ΔK2=0.2+0.3=0.5 N/m

  %Error in K=0.530×100=1.67%



Q 21 :

The time period of a simple harmonic oscillator is T=2πkm. Measured value of mass (m) of the object is 10 g with an accuracy of 10 mg and time for 50 oscillations of the spring is found to be 60 s using a watch of 2 s resolution. Percentage error in determination of spring constant (k) is ______ %.   [2026]

  • 3.35

     

  • 6.76

     

  • 7.60

     

  • 3.43

     

(2)

ΔKK=2ΔTT+Δmm

T=6050=1.2 sec

ΔT=250

 ΔKK=2×250×1.2+10×10-310=0.0676

 % Error=6.76%



Q 22 :

A spherical body of radius r and density σ falls freely through a viscous liquid having density ρ and viscosity η and attains a terminal velocity v0. Estimated maximum error in the quantity η is: (Ignore errors associated with σ, ρ and g,gravitational acceleration)   [2026]

  • 2[Δrr-Δv0v0]

     

  • 2[Δrr+Δv0v0]

     

  • 2Δrr-Δv0v0

     

  • 2Δrr+Δv0v0

     

(4)

V0=2r2g9n(ρB-ρL)

n=2r2g9V0(ρB-ρL)

Δnn=2Δrr+ΔV0V0



Q 23 :

If E and H represent the intensity of electric field and magnetising field respectively, then the unit of E/H will be    

  • ohm

     

  • newton

     

  • mho

     

  • joule

     

Select one or more options

If E and H represent the intensity of electric field and magnetising field respectively, then the unit of E/H will be    

write answer: ohm, newton & mho