A bucket containing water of depth of 10cm is kept in a lift which is vertically upwards with an acceleration which is eaqual to 1 1/2 times of acceleration due to gravity. Then pressure intensity at the bottom of the bucket in kg/sq.cm is
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Here we have use pseudo force concept.
means when bucket lift with acceleration a
then net acceleration act on bucket = (g+a)
now, density of water = 1g/cm³ = 1000 Kg/m³
mass = volume × density
= (c.s.a * depth) × 10^-3 Kg
= 1000 × 1/10 × A [ depth = 10cm and C.SA = A]
= 100A Kg
Fnet = m(g + a)
a = 3g/2 so, Fnet =m(g+3g/2)= m(5g/2) N
Fnet = 100A × 5 × 10/2 = 25 × 100A N
pressure = F/area
= 25 × 100A/A = 25 × 100 N/m²
= 2500 N/m²
= 250 kgwt/m²
now , pressure = 250 Kgwt/m²
your answer should be 250 Kgwt/m²
means when bucket lift with acceleration a
then net acceleration act on bucket = (g+a)
now, density of water = 1g/cm³ = 1000 Kg/m³
mass = volume × density
= (c.s.a * depth) × 10^-3 Kg
= 1000 × 1/10 × A [ depth = 10cm and C.SA = A]
= 100A Kg
Fnet = m(g + a)
a = 3g/2 so, Fnet =m(g+3g/2)= m(5g/2) N
Fnet = 100A × 5 × 10/2 = 25 × 100A N
pressure = F/area
= 25 × 100A/A = 25 × 100 N/m²
= 2500 N/m²
= 250 kgwt/m²
now , pressure = 250 Kgwt/m²
your answer should be 250 Kgwt/m²
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