1.The Gravitational unit of thrust in M.K.S system is
Ans.kgf
2.To reduce the pressure on a given area the total force applied is
Ans. To reduce/ decrease.
3.1 Pascal is equal to
Ans. 1N/m²
4.An Elephant walk on any kind of ground
Ans.surface area occupied by its feet is more.
5.A man of mass 75 kg is sleeping on a bed .if the the area of a leg of the bed in contact with the ground is 7.5 cm square. find the pressure on the floor exerted by that man
Ans.mass = 75kg
W=75 kg × g (W= m×g)
W= 75 Kgf
Area of leg of bed = 7.5 cm²
In a bed number of leg =4
So , area of 4 legs of bed= 4×7.5cm²
=30.0 cm²
P=Thrust/ Area
P=75 Kgf / 30cm²
= 7.5/3 kgfcm-²
= 2.5 kgfcm-²
= 2.5 × 10 N / 10-⁴ m². (1kgf= 9.8N ≈ 10N) & (1cm= 1/100 m ,1cm²= 10-⁴ m²)
= 2.5 ×10⁵ Nm-².
6.The thrust exerted by the blood in the veins our body is
Ans. Is slightly more than 1 atmospheric pressure.
7.Pressure exerted by a fluid is due to its
Ans weight.
8.Pressure inside a liquid increases with
Ans. Height of the liquid column.
9.To obtain a given moment of force for turning a body the force needed can be decreased by
Ans by applying force at the farthest point.
10.A person weighing 120 kgf stand on a platform of dimension 2.5 cm × 0.5 cm. what pressure in Pascal does he exert
Ans.W= 120 Kgf
W= 120× 10N (1kgf=9.8N ≈ 10N)
= 1200 N
Dimensions means, length=2.5 cm
Breadth= 0.5cm
Area=length×breadth
= 2.5/ 100 m × 0.5/100 m
= 1.25/10000 m²
P= thrust/area
= 1200N/( 1.25/10000 m²)
= 1200×10000/ 1.25 Nm-²
=960 ×10000 Nm-²
= 9.60×10²×10⁴ Nm-²
= 9.6 × 10⁶ pa (1Nm-² = 1Pa )
11.The moment of a force of of 50 Newton about a point is 2.5 Newton metre. find the perpendicular distance of force from that point
Ans. Moment of force= 2.5 Mm
force= 50N
Perpendicular distance=?
Moment of force=perpendicular distance×force applied
Moment of force= d × F
d=moment of force/force
= 2.5Nm/ 50N
= 0.25/5 m
= 0.5 m
= 50cm.
= 500 mm
12.1 kgf is equal to
Ans. 9.8 N
13.What is the magnitude of thrust required in Newton to produce a pressure of 26500 Pascal on an area of 100 cm square?
Ans. Pressure=26500 pa
Area=100 cm²
= 100 × 10-⁴ m²
= 10-² m²
Pressure=thrust/area
thrust= pressure× area
= 26500 Pa × 10-² m²
= 2.6500 ×10⁴ Nm-² × 10-² m² (1Pa = 1Nm-²)
= 2.65 ×10² N
14.A normal force of 100 Newton can produce a pressure of 100000 Pascal .calculate the area in centimetre square on which the force shall act to exert the pressure
Ans.F= 100N
P= 10⁵ Pa
A=?
P= F/A
A= F/ P
= 100 N/ 10⁵ Pa
= 10² N / 10⁵ Nm-²
= 10-³ m²
= 10-³ × 10⁴ cm² (1m²= 10⁴ cm²)
= 10¹ cm²
15.At sea level the thrust on an area of 6 square metre is equal to
Ans. 1atm=1.013× 10⁵ Nm-² (at sea level)
But,
P = Thrust / area
Thrust = P × area
= 1.013 × 10⁵ Pa × 6m²
= 6.078 × 10⁵ Nm-² × m²
= 6.078 × 10⁵ N
No comments:
Post a Comment
Thanks