Physics-
General
Easy

Question

A conical flask of mass 10 kg and base area as 103 cm2 is floating in liquid of relative density 1.2, as shown in the figure. The force that liquid exerts on curved surface of conical flask will be (Given g = 10 m/s2)

  1. 20 N downward direction    
  2. 40 N in downward direction    
  3. can be determined only if q is given    
  4. 20 N in upward direction    

The correct answer is: 20 N downward direction

Book A Free Demo

+91

Grade*

Related Questions to study

General
Physics-

A vertical jet of water coming out of a nozzle with velocity 20 m/s supports a plate of mass M stationary at a height h = 15m, as shown in the figure. If the rate of water flow is 1 litre per second, the mass of the plate is (Assume the collision to be inelastic).

Force by liquid = Mg

But and

A vertical jet of water coming out of a nozzle with velocity 20 m/s supports a plate of mass M stationary at a height h = 15m, as shown in the figure. If the rate of water flow is 1 litre per second, the mass of the plate is (Assume the collision to be inelastic).

Physics-General
Force by liquid = Mg

But and

General
physics-

A thin movable plate is separated from two fixed plates P subscript 1 and P subscript 2 by two highly viscous liquids of coefficients of viscosity n subscript 1 and n subscript 2 as shown, where n subscript 2 end subscript equals 9 n subscript 1 end subscript. Area of contact of movable plate with each fluid is same. If the distance between two fixed plates is ‘h’, then the distance ‘h subscript 1 end subscript’ of movable plate from upper plate such that movable plate can be moved with a finite velocity by applying the minimum possible force on movable plate is ( assume only linear velocity distribution in each liquid).

Viscous force due to upper liquid equals n subscript 1 end subscript A open parentheses fraction numerator v minus o over denominator h subscript 1 end subscript end fraction close parentheses
Viscous force due to lower liquid = n subscript 2 end subscript A open parentheses fraction numerator v minus o over denominator h minus h subscript 1 end subscript end fraction close parentheses
If total force is minimum
fraction numerator d over denominator d h subscript 1 end subscript end fraction open square brackets fraction numerator n subscript 1 end subscript over denominator h subscript 1 end subscript end fraction plus fraction numerator n subscript 2 end subscript over denominator h minus h subscript 1 end subscript end fraction close square brackets equals 0

A thin movable plate is separated from two fixed plates P subscript 1 and P subscript 2 by two highly viscous liquids of coefficients of viscosity n subscript 1 and n subscript 2 as shown, where n subscript 2 end subscript equals 9 n subscript 1 end subscript. Area of contact of movable plate with each fluid is same. If the distance between two fixed plates is ‘h’, then the distance ‘h subscript 1 end subscript’ of movable plate from upper plate such that movable plate can be moved with a finite velocity by applying the minimum possible force on movable plate is ( assume only linear velocity distribution in each liquid).

physics-General
Viscous force due to upper liquid equals n subscript 1 end subscript A open parentheses fraction numerator v minus o over denominator h subscript 1 end subscript end fraction close parentheses
Viscous force due to lower liquid = n subscript 2 end subscript A open parentheses fraction numerator v minus o over denominator h minus h subscript 1 end subscript end fraction close parentheses
If total force is minimum
fraction numerator d over denominator d h subscript 1 end subscript end fraction open square brackets fraction numerator n subscript 1 end subscript over denominator h subscript 1 end subscript end fraction plus fraction numerator n subscript 2 end subscript over denominator h minus h subscript 1 end subscript end fraction close square brackets equals 0
General
physics-

Figure shows a stream of fluid emerging from a tube in the base of an open fixed tank. The expression of ‘y’ (Maximum height traveled by jet of water) is

y equals fraction numerator u subscript y end subscript superscript 2 end superscript over denominator 2 g end fraction
u equals square root of 2 g h end root u subscript y end subscript equals square root of 2 g h end root sin invisible function application theta rightwards double arrow y equals fraction numerator open parentheses square root of 2 g h end root sin invisible function application theta close parentheses to the power of 2 end exponent over denominator 2 g end fraction equals h sin to the power of 2 end exponent invisible function application theta

Figure shows a stream of fluid emerging from a tube in the base of an open fixed tank. The expression of ‘y’ (Maximum height traveled by jet of water) is

physics-General
y equals fraction numerator u subscript y end subscript superscript 2 end superscript over denominator 2 g end fraction
u equals square root of 2 g h end root u subscript y end subscript equals square root of 2 g h end root sin invisible function application theta rightwards double arrow y equals fraction numerator open parentheses square root of 2 g h end root sin invisible function application theta close parentheses to the power of 2 end exponent over denominator 2 g end fraction equals h sin to the power of 2 end exponent invisible function application theta
General
physics-

For the arrangement shown in figure the time interval after which the water jet ceases to cross the wall (area of cross section of tank is A and orifice is ‘a’)

Velocity of efflux = square root of 2 g h end root
Find ‘h’ to have range of ejected water greater or equal than x
rightwards double arrow x equals square root of 2 g h end root. square root of fraction numerator 2 y over denominator g end fraction end root rightwards double arrow h equals fraction numerator x to the power of 2 end exponent over denominator 4 y to the power of 2 end exponent end fraction
time taken by liquid to drain out from H to h is equals fraction numerator A over denominator a end fraction square root of fraction numerator 2 over denominator g end fraction end root open square brackets square root of H minus square root of h close square brackets

For the arrangement shown in figure the time interval after which the water jet ceases to cross the wall (area of cross section of tank is A and orifice is ‘a’)

physics-General
Velocity of efflux = square root of 2 g h end root
Find ‘h’ to have range of ejected water greater or equal than x
rightwards double arrow x equals square root of 2 g h end root. square root of fraction numerator 2 y over denominator g end fraction end root rightwards double arrow h equals fraction numerator x to the power of 2 end exponent over denominator 4 y to the power of 2 end exponent end fraction
time taken by liquid to drain out from H to h is equals fraction numerator A over denominator a end fraction square root of fraction numerator 2 over denominator g end fraction end root open square brackets square root of H minus square root of h close square brackets
General
physics

The distance travelled by a particle is given by S=3+2t +5t2The initial velocity of the particle is…..

The distance travelled by a particle is given by S=3+2t +5t2The initial velocity of the particle is…..

physicsGeneral
General
maths-

If f colon R not stretchy rightwards arrow R text  defined by  end text f left parenthesis x right parenthesis equals x squared minus 2 x minus 3 text  then  end text f text  is  end text

If f colon R not stretchy rightwards arrow R text  defined by  end text f left parenthesis x right parenthesis equals x squared minus 2 x minus 3 text  then  end text f text  is  end text

maths-General
General
physics

A freely falling particle covers a building of 45 m height in one second. Find the height of the point from where the particle was released.[g=10 ms-2]

A freely falling particle covers a building of 45 m height in one second. Find the height of the point from where the particle was released.[g=10 ms-2]

physicsGeneral
General
physics

The displacement of a particle in x direction is given by x.=9-5t+4t2 Find the Velocity at time t=0

The displacement of a particle in x direction is given by x.=9-5t+4t2 Find the Velocity at time t=0

physicsGeneral
General
Maths-

If f colon left square bracket 0 comma 1 right square bracket not stretchy rightwards arrow left square bracket negative 1 comma 3 right square bracket defined by f left parenthesis x right parenthesis equals x squared plus x plus 1 text , then  end text f text  is  end text

If f colon left square bracket 0 comma 1 right square bracket not stretchy rightwards arrow left square bracket negative 1 comma 3 right square bracket defined by f left parenthesis x right parenthesis equals x squared plus x plus 1 text , then  end text f text  is  end text

Maths-General
General
Maths-

If f left parenthesis x right parenthesis equals vertical line x minus 1 vertical line plus vertical line x minus 2 vertical line plus vertical line x minus 3 vertical line comma f colon left square bracket 2 comma 3 right square bracket not stretchy rightwards arrow R text  is  end text

If f left parenthesis x right parenthesis equals vertical line x minus 1 vertical line plus vertical line x minus 2 vertical line plus vertical line x minus 3 vertical line comma f colon left square bracket 2 comma 3 right square bracket not stretchy rightwards arrow R text  is  end text

Maths-General
General
physics-

Two thin metallic strips, carrying current in the direction shown, cross each other perpendicularly without touching but being close to each other, as shown in the figure. The regions which contain some points of zero magnetic induction are

) II and IV

Hence, magnetic induction in region I and IV will be zero.

Two thin metallic strips, carrying current in the direction shown, cross each other perpendicularly without touching but being close to each other, as shown in the figure. The regions which contain some points of zero magnetic induction are

physics-General
) II and IV

Hence, magnetic induction in region I and IV will be zero.
General
maths-

A equals left curly bracket x colon negative 1 less or equal than x less or equal than 1 right curly bracket. f colon A not stretchy rightwards arrow A defined by f left parenthesis x right parenthesis equals x vertical line x vertical linetext  Then  end text f text  is  end text

A equals left curly bracket x colon negative 1 less or equal than x less or equal than 1 right curly bracket. f colon A not stretchy rightwards arrow A defined by f left parenthesis x right parenthesis equals x vertical line x vertical linetext  Then  end text f text  is  end text

maths-General
General
physics

What does the speedometer measure kept in motorbike ?

What does the speedometer measure kept in motorbike ?

physicsGeneral
General
physics-

A cell is connected between the points A and C of a circular conductor ABCD with O as centre and angle A O C equals 60 degree. If B subscript 1 and B subscript 2 are the magnitudes of the magnetic fields at O due to the currents in  ABC and ADC respectively, then ratio fraction numerator B subscript 1 end subscript over denominator B subscript 2 end subscript end fraction is

From Biot-Savart law the magnetic field at the centre is directly proportional to the length of current carrying segment.
therefore fraction numerator B subscript 1 end subscript over denominator B subscript 2 end subscript end fraction equals fraction numerator l e n g t h blank o f blank A B C over denominator l e n g t h blank o f blank A D C end fraction
equals fraction numerator a n g l e blank s u b t e n d e d blank b y blank A B C over denominator a n g l e blank s u b t e n d e d blank b y blank A D C end fraction
equals fraction numerator left parenthesis 360 degree minus 60 degree right parenthesis over denominator 60 degree end fraction equals fraction numerator 300 over denominator 60 end fraction equals fraction numerator 5 over denominator 1 end fraction

A cell is connected between the points A and C of a circular conductor ABCD with O as centre and angle A O C equals 60 degree. If B subscript 1 and B subscript 2 are the magnitudes of the magnetic fields at O due to the currents in  ABC and ADC respectively, then ratio fraction numerator B subscript 1 end subscript over denominator B subscript 2 end subscript end fraction is

physics-General
From Biot-Savart law the magnetic field at the centre is directly proportional to the length of current carrying segment.
therefore fraction numerator B subscript 1 end subscript over denominator B subscript 2 end subscript end fraction equals fraction numerator l e n g t h blank o f blank A B C over denominator l e n g t h blank o f blank A D C end fraction
equals fraction numerator a n g l e blank s u b t e n d e d blank b y blank A B C over denominator a n g l e blank s u b t e n d e d blank b y blank A D C end fraction
equals fraction numerator left parenthesis 360 degree minus 60 degree right parenthesis over denominator 60 degree end fraction equals fraction numerator 300 over denominator 60 end fraction equals fraction numerator 5 over denominator 1 end fraction
General
physics-

Current I is flowing in conductor shaped as shown in the figure. The radius of the curved part is r and the length of straight portion is very large. The value of the magnetic field at the centre O will be

B subscript A end subscript equals 0

B subscript B end subscript equals blank fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator left parenthesis 2 pi minus pi divided by 2 right parenthesis I over denominator r end fraction blank circled times equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator 3 pi I over denominator 2 r end fraction
B subscript C end subscript equals fraction numerator mu subscript 0 end subscript I over denominator 4 pi r end fraction circled times
So, net magnetic field at the centre
equals B subscript A end subscript plus B subscript B end subscript plus blank B subscript C end subscript
equals 0 plus fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator 3 pi I over denominator 2 r end fraction plus fraction numerator mu subscript 0 end subscript I over denominator 4 pi r end fraction equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator I over denominator r end fraction open parentheses fraction numerator 3 pi over denominator 2 end fraction plus 1 close parentheses

Current I is flowing in conductor shaped as shown in the figure. The radius of the curved part is r and the length of straight portion is very large. The value of the magnetic field at the centre O will be

physics-General
B subscript A end subscript equals 0

B subscript B end subscript equals blank fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator left parenthesis 2 pi minus pi divided by 2 right parenthesis I over denominator r end fraction blank circled times equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator 3 pi I over denominator 2 r end fraction
B subscript C end subscript equals fraction numerator mu subscript 0 end subscript I over denominator 4 pi r end fraction circled times
So, net magnetic field at the centre
equals B subscript A end subscript plus B subscript B end subscript plus blank B subscript C end subscript
equals 0 plus fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator 3 pi I over denominator 2 r end fraction plus fraction numerator mu subscript 0 end subscript I over denominator 4 pi r end fraction equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator I over denominator r end fraction open parentheses fraction numerator 3 pi over denominator 2 end fraction plus 1 close parentheses