Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
two identical pith balls are charged by rubbing against each other.they are suspended from a horizontal rod through two strings of length 20cm each, the separation between the suspension points being 5 cm.in equilibrium, the separation between the balls is 3 cm find the mass of each ball and tension in the strings..the charge on each ball has a magnitude 2×10^-8 cm? plz, can u explain each step..plz....cz I need full step ......
Let θ is the angle made by string with vertical . Here tanθ = perpendicular/base = √{20² - 1²}/1 = √{399} ≈ 20 e.g., cosθ = base/perpendicular = 20{approximately}/20 = 1 Hence, cosθ = 1 and sinθ = perpendicular/hypotenuse= 1/20 Now, at equilibrium Tcosθ = mg { weight } -----(1) Tsinθ = F { electrostatic force between them } ------(2)Now, F = kq₁q₂/r² Here , q₁ = q₂ = 2 × 10⁻⁸ C and r = 3cm = 0.03 m Now, F = 9 × 10⁹ × 2 × 10⁻⁸ × 2 × 10⁻⁸/(0.03)² = 9 × 10⁹ × 4 × 10⁻¹⁶/9 × 10⁻⁴ = 4 × 10⁻³ N Now, Tsinθ = 4 × 10⁻³ or, T × 1/20 = 4 × 10⁻³ [ sinθ = 1/20 }or, T = 4 × 20 × 10⁻³ = 0.08N Hence, tension in string is 0.08N Now, Tcosθ = mgor, T × 1 = m × 10 [ g = 10 m/s²or, 0.08 = m × 10 ⇒ m = 0.008 Kg Hence, mass of each ball is 8g
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -