Numerical on Significant Figures Class 11 Physics ISC Nootan Solutions Ch-2 Unit and Measurement. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-11 Physics.
Numerical on Significant Figures Class 11 Physics ISC Nootan Solutions Ch-2 Unit and Measurement
| Board | ISC |
| Class | 11 |
| Subject | Physics |
| Book | Nootan |
| Chapter-2 | Unit and Measurement |
| Topics | Numerical on Significant Figures |
| Academic Session | 2024-2025 |
Numerical on Significant Figures
( Class 11 Physics ISC Nootan Solutions Ch-2 Unit and Measurement)
Que-29: Add the following with due consideration of significant figures :
(i) 3.8*10^-8 + 4.2*10^-6.
(ii) 3.8*10^-7 +4.2*10^-6.
(iii) 4.22*10^5 + 3.11*10^7 + 6.003*10^6
(iv) 1.294 cm + 35.1 cm.
(v) 348.4 km + 105 km + 143.8 km.
(vi) 1.5 kg + 264g + 52mg.
Sol: (i) 3.8×10^-8 + 4.2×10^-6
=.038×10^-6 + 4.2×10^-6
= 4.238×10^-6
up to two significant figure
= 4.2×10^-6 Ans.
(ii) 3.8×10^-7 + 4.2×10^-6
= 0.38×10^-6 + 4.2×10^-6
= 4.58×10^-6
up to two significant figure
= 4.6×10^-6 Ans.چھ
(iii) 4.22×10^5 + 3.11×10^7 + 6.003×10^6
.0422×10^7 + 3.11×10^7 + 0.6003×10^7
= 3.7525×10^7
up to three significant figure
= 3.75×10^7 Ans.
(iv) 1.294 cm + 35.1 cm = 36.394
up to three significant figure
= 36.4 Ans.
(v) 348.2 km + 105 km + 1438 km
= 597.0 km
up to 3 significant figure
= 597 Ans.
(vi) 1.5 + 0.264 + 0.00052
= 1.8 kg Ans.
Que-30: Subtract the following with due consideration of significant figures :
(i) (5.0*10^-4) – (2.5*10^-6)
(ii) (18.4 cm) – (18.132 cm)
(iii) (104.4g) – (2.34g)
(iv) (4.0*10^2 kg) – (41.72 kg)
(v) (172.4 kg) – (98.767 kg)
Sol: (i) 5 x 10^-4
(ii) 0.3 cm
(iii) 102.1 g
(iv) 3.6 x 10² kg
(v) 172.3 kg
Que-31: A jeweller puts a diamond weighing 4.31g in a box weighing 1.5kg. Find the total weight up to appropriate number of significant number.
Sol: weight of diamond = 4.31 kg
So, g = 0.00431 kg
Then, total weight = 1.5 + 0.00431
= 1.50542 kg
Hence, the combined mass of diamond and box is 1.5 kg Ans.
Que-32: Write the value of the product 185 x 1.52 in significant figures.
Sol: 185×1.52 = 281.2
It has four significant figures = 281.
Que-33: The side of the square is 1.6m. Write its area in appropriate significant figures.
Sol: side of square = 1.6m
area of square = side ^2
= (1.6)^2
= 2.56m^2
ans. = 2.6m^2
Que-34: The length of path-strip is 10.53m and its width is 0.97m. Compute its area in appropriate significant figures.
Sol: Length of path-strip = 10.53m
Width of path-strip = 0.97m
Area of path-strip = length x breadth
Area of path-strip = 10.53 x 0.97
= 10.241 m^2 = 10m^2 Ans.
Que-35: The length, breadth and height of a block are 12.1cm, 6.3cm and 8.4mm. Find its volume in appropriate significant figures.
Sol: length of block (l) = 12.1cm
breadth of block (b) = 6.3cm
height of block (h) = 8.4mm = 0.84cm
volume of block = l x b x h
volume of block =12.1 x 6.3 x 0.84
= 64.03 cm^3 = 64 cm^3 Ans.
Que-36: Find the volume, in significant figures, of the block of length 20m, width 25cm and thickness 13.53cm.
Sol: length of block (l) = 20m
width of block (b) = 25cm = 0.25m
thickness of block (h) = 13.53cm = 0.1353m
Volume of block = l x b x h
Volume of block = 20 x 0.25 x 0.1353
=0.6765 cm^3 = 0.68 m^3 Ans.
Que-37: A circle has a diameter of 5.2cm. Write its circumference in significant figures.
Sol: Circumference = 2πr = (2π) d/2
(2×3.14×5.2)/2 = 16.328cm
answer up to two significant figure = 16cm.
Que-38: The diameter of sphere is 4.24cm. Compute its surface are up to appropriate significant figures.
Sol: Surface area = 4πr^2
4π x (d/2)^2 = πd^2
= 3.142×(4.24)² cm²
56.495 cm²
Answer upto three significant figure.
56.5 cm² Ans.
Que-39: A cylinder has a length of 1.0×10^-1 m and diameter of 2.40×10^-3 m. Find the cross-sectional area and volume of cylinder with due consideration of significant figures.
Sol: Cross section area = πr²= (πd^2)/4
3.14 x (2.40×10^-3)^2 = 4.52×10^-6 m² Ans.
Volume of cylinder
=(πhd^2)/4
= 4.52×10^-6×1.0x10^-1
= 4.5 x 10^-7 m³ Ans.
Que-40: A thin rectangular leaf has a surface area of 3.41m^2 and width of 1.034m. Find its length of correct number of significant figures.
Sol: Length = area/width
=3.41m²/1.034m
=3.30m Ans.
Que-41: The volume of 91.3g of ball is 25cm^3. Find its density in correct number of significant figures.
Sol: P = M/V
= 91.3gm/25 cm^-3
= 3.6 gm cm^-3 Ans.
Que-42: A body moves 1250m in 11s. Find its average speed in correct number of significant figures.
Sol: Average speed = distance/time
1250/11 = 1.1×10^2 m/s Ans.
Que-43: Solve (2.91×0.3842)/0.080 with due consideration of significant figures.
Sol: (2.91 x 0.3842)/0.080 = 14
answer in a 2 digits because out of all operating number minimum significant no is 2 (0.080)
Que-44: In a simple pendulum experiment, the measured length of pendulum is 90.6×10^-2 m and the period of oscillation is 1.91 s. Find the value of acceleration due to gravity up to correct significant figures.
Sol: by formula. T=2π√l/g
g = (4π^2l)/T^2
g = {4 x (3.14)^2 x 90.6 x 10^-2}/1.91
g = 9.81m s^-2
by rule of significant figures.
Que-45: The measured mass and diameter of a uniform brass ball are 29.150×10^-3 kg and 1.92×10^-2 m respectively. Express the density of brass in appropriate significant figures.
Sol: – P = M/V
P = {29.150 x 10^-3 x 8}/ 4/3 x 3.14 x (1.92/2 x 10^-2)^3
= {3 x 29.150×10^-3 x 8}/[4 x 3.14 x (1.92 x 10^-2)^3]
= 7.87 x 10^3 kg m^-3
by the rules of significant figure
— : end of Numerical on Significant Figures Class 11 Physics ISC Nootan Solutions Ch-2 :–
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