# Error Analysis, Percentage Error

***Numerical questions***

**1. The width of a road of 20 meters is measured with the help of a 1 meter long rod. What is the maximum percentage error that can occur from this measurement?**

**Solution-**

x=20 m

∆x=1 m

percent error,

=(∆x/x)×100

=(1/20)×100

=5%

**2. The length of a solid is measured to be 7.50 cm with a vernier callipers. If the minimum scale of the vernier is 0.01 cm, calculate the possible percentage error of the appropriate measurement.**

**Solution-**

l=7.50 cm

∆l=0.01 cm

percent error,

=(∆l/l)×100

=(0.01/7.50)×100

=(10/75)

=0.134%

**3. The least count of a screw gauge is 0.01 mm. Find the percentage error in measuring 2 mm thickness with the help of this instrument.**

**Solution-**

r=2 mm

∆r=0.01 mm

percent error,

=(∆r/r)×100

=(0.01/2)×100

=0.50%

**4. The least count of a screw gauge is 0.001 cm. By this the diameter of a wire is measured to be 0.225 cm. Calculate possible percentage error in this measurement.**

**Solution-**

r=0.225 cm

∆r=0.001 mm

percent error,

=(∆r/r)×100

=(0.001/0.225)×100

=0.445%

**5. If there is an error of 1.5% in measuring the radius of a circle, then what will be the maximum percentage error in finding its area.**

**Solution-**

(∆r/r)×100=1.5% (given)

area of circle,

A=πr²

A=2×(∆r/r)×100

A=2×1.5

A=3%

**6. If there is an error of 0.5% in measuring the radius of a sphere, calculate the maximum percentage error in finding its volume.**

**Solution-**

Volume of sphere,

V=(4/3)πr³

(∆r/r)×100=0.5% (given)

(∆v/v)×100=3×(∆r/r)×100

(∆v/v)×100=3×0.5

(∆v/v)×100=1.5%

**7. If there is an error of 1.5% in measuring the side of a cube, then find the maximum percentage error in finding the volume of the cube.**

**Solution-**

Volume of cube,

V=a³

(∆a/a)×100=1.5% (given)

(∆v/v)×100=3×(∆a/a)×100

(∆v/v)×100=3×1.5

(∆v/v)×100=4.5%

**8. If the error in measuring the length, breadth and height of a cuboid is 2%, 3% and 4% respectively, then find the total possible error in the volume.**

**Solution-**

∆l/l×100=2%

∆b/b×100=3%

∆h/h×100=4%

Total possible error in volume,

∆v/v×100=(∆l/l)×100+(∆b/b)×100+(∆h/h)×100

(∆v/v)×100=2%+3%+4%

(∆v/v)×100=9%

**9. The least count of the stopwatch is 1/5 second. Due to this, the time taken for 20 oscillations of a pendulum is measured as 25 seconds. Find the possible percentage error in this observation.**

**Solution-**

∆t=1/5 sec

t=25 sec

percent error,

=(∆t/t)×100

=(1/5)/25

=100/125

=0.8%

**10.**The length of a rectangular sheet is measured to be 5.7 cm and width to be 3.4 cm using an ordinary scale giving measurements up to 0.1 cm. What is the percentage error in the area of the sheet.

**Solution-**

[A=l.b]

∆l=0.1 cm

∆b=0.1 cm

Percentage error in area,

(∆A/A)×100=(∆l/l)×100+(∆b/b)×100

(∆A/A)×100=(0.1/5.7)×100+(0.1/3.4)×100

(∆A/A)×100=(100/57)+(100/34)

(∆A/A)×100=1.75+2.94

(∆A/A)×100=4.69%

**11.**

**The following observations were obtained in determining the surface tension of water through a capillary tube, diameter of the capillary tube D=0.125cm and elevation of water in the capillary h=1.43cm, maximum possible percentage in the value of surface tension T by using the formula T=rhg/2. Find the error.**

**Solution-**

D=2r=0.125cm

h=1.43cm

T=rhg/2

T=2rhg/4

T=Dhg/4

(∆T/T)×100=(∆D/D)×100+(∆h/h)×100

(∆T/T)×100=(0.001)/(0.125)×100+(0.01)/(1.43)×100

(∆T/T)×100=(100/125)+(100/143)

(∆T/T)×100=1.49%

12.

**The speed of sound in a taut string is v=√(T/m), where T=Mg. Find the maximum possible per cent error in the value of v using M=2.0 kg and m=1.5 g/m.****Solution-**

v=√(T/m)

T=Mg

v=√(Mg/m)

M=2.0kg

m=1.5 g/m

(∆v/v)×100=1/2(∆M/M)×100+1/2(∆m/m)×100

=1/2(0.1/2.0)×100+1/2(0.1/1.5)×100

=1/2(100/20)+1/2(100/15)

=5/2+50/15

=175/30

(∆v/v)×100=5.8%

**13. In an experiment the values a=2.154, b=2.0, c=0.368 were obtained. The formula Z=a³bc⁻² is used experimentally. Find the maximum possible percentage error in the value of Z.**

**Solution-**

a=2.154

b=2.0

c=0.368

Z=a³bc⁻²

(∆Z/Z)×100=3(∆a/a)×100+(∆b/b)×100+2(∆c/c)×100

=3(.001/2.153)×100+(0.1/2.0)×100+(0.001/0.368)×100

=(100/718)+5+(100/184)

=0.13+5+0.54

(∆Z/Z)×100=5.67%

**14. Two rods whose lengths measure (0.3±0.1) meter and (4.1±0.2) meter are joined together. Find the maximum percentage error in the combined length of the rod.**

**Solution-**

combined length of both rods,

l=l₁+l₂

l=4.1+3.5

l=7.6 meters

∆l=0.2+0.1

∆l=0.3 meter

percent error,

(∆l/l)×100=(0.3)/(7.6)×100

(∆l/l)×100=300/76

(∆l/l)×100=3.9%

**15. The radius of a sphere is (5.3±0.1) cm. Find maximum percentage error in the volume of the sphere.**

**Solution-**

volume of sphere,

[V=4/3πr³]

r=5.3

∆r=0.1

Percentage error in volume,

(∆v/v)×100=3(∆r/r)×100

(∆v/v)×100=3×(0.1/5.3)×100

(∆v/v)×100=300/53

(∆v/v)×100=5.67%

**16. A physical quantity S is related to three measured quantities a, b and c in the following way:**

S=a³b²/c

The errors in the measurements of a,b,c are 1%, 2%, 3% respectively. Find the maximum possible percentage error in the value of quantity S.

**Solution-**

(∆s/s)×100=3(∆a/a)×100+2(∆b/b)×100-(∆c/c)×100

(∆s/s)×100=3×1+2×2-3

(∆s/s)×100=3+4-3

(∆s/s)×100=4%

**17. The density of a substance is 8.4 g/cm³ whereas experimentally it is found to be 8.7 g/cm³. Find the percentage error.**

**Solution-**

d=8.4 g/cm³

d´=8.7 g/cm³

Experimental percentage error = {(experimental value – actual value)/actual value}×100

Experimental percentage error={(8.7-8.4)÷8.4}×100

Experimental percentage error=(0.3/8.4)×100

Experimental percentage error=300/84

Experimental percentage error=3.57%