Adayar – Adambakkam – Pallavaram – Pammal – Chromepet – Selaiyur
+1th PHYSICS
(Syllabus: Nature of the physical world and measurement)
I.
Numerical problems and solutions:-
1.
How many astronomical units are there in 1 metre?
Given data:- One astronomical unit (AU) = 1.496 x 1011 m
Solution:- number of astronomical units in 1 metre = 1/1.496 x 1011
= 0.6684 x 10-11
=6.684x 10I2AU.
2.
If mass of an electron is 9.11 x 10-31 kg, how many electrons would
weigh 1 kg?
Given Data:- Mass of an electron = 9.11 x 10 - 3 1 kg
Solution:- number of electrons in 1 kg = 1/9.11 x 10 - 3 1
= 0.1097 x 1031
= 1.097 x 1030 electrons
3.
In a submarine emitted with a SONAR, the time delay between
generation of a signal and reception of its echo after reflection from
an enemy ship is observed to be 73.0 seconds. If the speed of sound
in water is 1450 ms-1. then calculate the distance of the enemy
ship.
Given Data:- Time delay in SONAR between generation of a signal and reception
of its echo after reflection from enemy ship
Speed of sound in water
= 73
= 1450 ms_l
Solution:- distance of the enemy ship = ( speed x time)/2
= (1450 x 73)/2 = (105850)/2.
= 52.925 Km.
4.
State the number of significant figures in the following.
i) 600900
ii) 5212.0
iii) 0.0631
iv) 0.0631
Solution:- i) 600900-number of significant figures = 4.
v)2.64x1024
ii) 5212.0-number of significant figures = 5.
iii) 6.320-number of significant figures = 4.
iv) 0.0631-number of significant figures = 3.
v) 2.64 x 1024-number of "significant figures = 3.
5.
Find the value of π2 correct to significant figures, if π = 3.14.
Solution:- the value of π2 = 9.8596
The correct significant figures are 9.86
6.
5.74 g of a substance occupies a volume of 1.2 cm3. Calculate its density
applying the principle of significant figures
Given data:- quantity of a substance = 5.74 g
Occupied volume = 1.2 cm2
Solution:- density = mass/ volume = 5.74/1.2 = 4.78 g/cm3.
7.
The length, breadth and thickness of a rectangular plate are 4.234
m, 1.005 m and 2.01 cm respectively. Find the total area and
volume of the plate to correct significant figures.
Given data:- length of the rectangular plate = 4.234 m
Breadth of the rectangular plate = 1.005 m
Thickness of the rectangular plate = 2.01 cm = 0.0201 m
Solution:- Total area = L x B = 4.234 x 1.005
Total area correct to significant figures = 4.255 m2.
Volume = L x b x h = 4.234 x 1.005 x 0.0201
Volume correct to significant figures = 0.0855 m3.
8.
The length of a rod is measured as 25 cm using a scale having an
accuracy of 0.1 cm. Determine the percentage error in length.
Given data:- Length of a rod measured using a scale = 25 cm
Accuracy of the scale = 0.1 cm
Solution:- percentage error in length = (ΔX x 100 %)/X
= 0.1/25 x 100 = 0.4 %.
9.
Obtain by dimensional analysis an expression for a surface tension
of a liquid rising in a capillary tube. Assume that the surface
tension T depends on mass m of the liquid, pressure P of the liquid
and radius r of the capillary tube. K = ½
Solution:- T = K mx py rz
m = M; p = ML-1T-2; r = L; F = MLT-2; l = L
Sub the value in the above equation ML0T-2 = KMx+y L- y+z T-2y
Comparing the equation x = 0; y = 1; z = 1; K = ½
T = ½ m 0p 1r 1
T = pr/2.
10.
The force F acting on a body moving in a circular path depends on
the mass m of the body, velocity v radius r of the circular path.
Obtain an expression for the force by dimensional analysis K = 1.
Solution:- F = K mx vy rz
m = M; v = LT-1; r = L; a = LT-2
sub the F = ma and other values in above equation
MLT-2 = K Mx Lx + y T2y
X = 1; y = 2; z = -1; K = 1
F = 1 m1 v2 r-1 = mv2/r.
11.
Check the correctness of the following equation by dimensional
analysis
i) F = mv2/r2 where F is force, m is mass, v is velocity and r is
radius
F = MLT-2 ; m = M; v = LT-1; r = L
Sub the values
MLT-2 = M(LT-1)2/L2 = MT-2.
ii) n = 1/2π√𝑔/𝑙 where n is frequency, g is acceleration due to
gravity and l is length.
n = T-1; g = LT-2; l = L
T-1 = 1/2π √𝐿𝑇-2/L = 1/2π T-1.
iii) ½ mv2 = mgh2 where m is mass, v is velocity, g is acceleration
due to gravity and h is height.
m = M; v = LT-1; g = LT-2; h = L
½ ML2T-2 = ML3T-2 Hence dimensionally incorrect
12.
Convert using dimensional analysis
i) 18/5 Kmph into ms-1
The dimensional formula for velocity v in Kmph = M1x L1y T1z
The dimensional formula for velocity v in mps = M2x L2y T2z
Comparing the coeff x = 0; y = 1; z = -1
Vmps = Vkmps (M1/M2)x (L1/L2)y (T1/T2)z
= 18/5 (1 Kg/1 g)0 (1 Km/1m)1 (1 hr/1 s)-1
= 3600 (3600)-1
=18/5 Km hr-1 = 1 ms-1.