The radius of the orifice of a liquid dropper is 2 mm. If the surface tension of water is
70 x 10^-3 N/m and g = 10 m/sec^2 then the mass of the water drop falling from dropper is
1) 8.8 x 10^-5 kg 2) 8.8 x 10^-4 kg 3) 8.8 x 10^-3 kg 4) 8.8 x 10^-2 kg
Answers
Answer :
- The mass of the water drop falling from the dropper = 8.8 × 10⁻⁵ kg
Explanation :
Given :
- Surface tension of the water drop, T = 70 × 10⁻³ N m⁻¹
- Acceleration due to gravity, g = 10 m s⁻²
- Radius of the drop, r = 2 mm or 0.002 m
To find :
- The mass of the water drop, m = ?
Knowledge required :
- Formula for surface tension :
⠀⠀⠀⠀⠀⠀⠀⠀⠀T = F/L
Where :
⠀⠀⠀⠀⠀⠀⠀➝ T = Surface tension
⠀⠀⠀⠀⠀⠀⠀➝ F = Force
⠀⠀⠀⠀⠀⠀⠀➝ L = Length
- Circumference of a circle :
⠀⠀⠀⠀⠀⠀⠀⠀⠀C = 2πr
[Where, r is the radius of the circle]
- Formula for force :
⠀⠀⠀⠀⠀⠀⠀⠀⠀F = ma
Where,
⠀⠀⠀⠀⠀⠀⠀➝ m = Mass
⠀⠀⠀⠀⠀⠀⠀➝ F = Force
⠀⠀⠀⠀⠀⠀⠀➝ a = Acceleration
[Note : Here, the length (L) on which the force exerts is the circumference of the drop of water.]
Solution :
First let us find the length on which the force acts.
By using the formula for circumference of a circumference of a circle and substituting the values in it, we get :
⠀⠀=> C = 2πr
⠀⠀=> L = 2 × 22/7 × 0.002 [Taking, π = 22/7]
⠀⠀=> L = 2 × 22/7 × 2/1000
⠀⠀=> L = 2 × 22/7 × 2 × 10⁻³
⠀⠀=> L = 88/7 × 10⁻³
⠀⠀⠀⠀∴ Length on which the force acts, L = 88/7 × 10⁻³ m.
Now let us find the force exerted on the drop of water :
By using the equation for surface tension and substituting the values in it, we get :
⠀⠀=> T = F/L
⠀⠀=> 70 × 10⁻³ = F/(88/7 × 10⁻³)
⠀⠀=> 70 × 10⁻³ = F/88 × 7 × 10⁻³
⠀⠀=> 70 × 10⁻³ = (7 × 10⁻³)F/88
⠀⠀=> 70 × 10⁻³ × 88 = (7 × 10⁻³)F
⠀⠀=> (70 × 10⁻³ × 88)/(7 × 10⁻³) = F
⠀⠀=> (10⁻³ × 88)/(10⁻²) = F
⠀⠀=> 10⁻⁵ × 88 = F
⠀⠀⠀⠀∴ Force exerted on the water drop, F = 88 × 10⁻⁵ N
Let's find the mass of the water drop now,
By using the formula for force and substituting the values in it, we get :
⠀⠀=> F = ma
⠀⠀=> 88 × 10⁻⁵ = m × 10
⠀⠀=> (88 × 10⁻⁵)/10 = m
⠀⠀=> 88 × 10⁻⁶ = m
⠀⠀=> 8.8 × 10⁻⁵ = m
Therefore,
- The mass of the water drop, m = 8.8.× 10⁻⁵ kg.
Answer:
Answer :
The maximum hieght reached by the ball is 45 m.
Explanation :
Given :
Angle of projection, θ = 30°
Intial velocity of the ball, u = 60 m/s
Final velocity of the ball, v = 0 m/s
[Final velocity of a body at maximum height is 0, v = 0]
Acceleration due to gravity, g = 10 m/s² (Approx.)
[Acceleration due to gravity = 9.8 m/s²]
To find :
Maximum height reached by the ball, h = ?
Knowledge required :
Formula for maximum height in case of a projectile :
\boxed{\sf{h_{(max.)} = \dfrac{u^{2}sin^{2}\theta}{2g}}}
h
(max.)
=
2g
u
2
sin
2
θ
Where,
h = Maximum height
u = Initial velocity
θ = Angle of Projection
g = Acceleration due to gravity
Solution :
By using the formula for maximum height in case of a projectile and substituting the values in it, we get :
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{u^{2}sin^{2}\theta}{2g}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
2g
u
2
sin
2
θ
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{60^{2} \times sin^{2}30^{\circ}}{2 \times 10}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
2×10
60
2
×sin
2
30
∘
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{60^{2} \times \bigg(\dfrac{1}{2}\bigg)^{2}}{20}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
20
60
2
×(
2
1
)
2
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{60^{2} \times \dfrac{1}{4}}{20}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
20
60
2
×
4
1
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{3600 \times \dfrac{1}{4}}{20}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
20
3600×
4
1
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{3600}{20 \times 4}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
20×4
3600
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{3600}{80}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
80
3600
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{360}{8}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
8
360
\begin{gathered}:\implies \sf{h_{(max.)} = 45} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=45
\begin{gathered}\boxed{\therefore \sf{h_{(max.)} = 45\:m}} \\ \\ \\ \end{gathered}
∴h
(max.)
=45m
Therefore,
Maximum height reached by the ball, h = 45 m.