Skip to main content

9th Physics Solved Numerical Floating bodies ( Gravitation) Term-2

Q. a cube of mass 1kg with each side of 1cm is lying on the table. find the pressure exerted by the block on the table. take g=10 m/s2

Ans: Pressure is given as force/area 

so, Force, F = mg = 1000 X 1000 gm.cm/s2

and   area, A = 1x1 cm2 = 1 cm2


Thus, the pressure exerted would be

P = (1000 X 1000) / 1 or P = 1 x 10 pa

Q. The mass of a solid iron cube of side 3cm is to be determined usig a spring balance. If the of iron is approximately 8.5 g/cm3, the best suited spring balance for determining weight of the solid would be of 
1. range 0-250gwt ; least count 1gwt             
2. range 0-250gwt ; least count 5gwt
3. range 0-1000gwt ; least count 5gwt           
4.  range 0-1000gwt ; least count 10gwt

Ans: Edge=3 cm ,  Density=8.5 g/cm3


Mass=  density x volume = 8.5 x(3x3)=229.5gwt

Therefore second spring balance of range 0-250 gwt with least count 5gwt will be suitable.

Q. The density of turpentine oil is 840 Kg/ m3. What will be its relative density. (Density of water at 4 degree C is 10 cube kg minus cube)


Ans:Relative Density = Density of Substance/ Density of water at 4 0c

Density of turpentine oil = 840 kg/ m3 ( given).

Density of water at 4 0c = 1000 kg/ m3Relative density of turpentine oil = Density of turpentine oil / 

Density of water at 4 0c  = (840 / 1000 ) kg m-3/ kg m-3     = 0.84

Since, the relative density of the turpentine oil is less than 1, therefore it will float in water.

Q. A solid body of mass 150 g and volume 250cm3 is put in water . will following substance float or sink if the density of water is 1 gm-3?

Ans: The substance will float if its density is less than water and will sink if its greater.

so, density of solid body is   d = mass/volume 

or d = 150/250 = 0.6 gm/cm3     which is less than the density of water (1 gm/3).

So, the solid body will float on water.

Q.  A body weighs 50 N in air and when immersed in water it weighs only 40 N. Find its relative density.


Ans: the relative density would be ratio of the density of the body with respect to air and the density of the body with respect to water.

so,  F1 = 50 N   F2 = 40 Nso,  F1/F2 = 50/40  or relative masses m1/m2 = 5/4 anddensity = mass/volume and as volume remains constant, Relative density =  d1/d2 = 5/4

Q.  A ball of relative density 0.8 falls into water from a height of 2m.  find the depth to which the ball will sink ?


Ans: Speed of the ball

V  = Ö2gh    = Ö 2x10x2  = 6.32 m/s

Buoyancy force by water try to stop the ball.

Buoyancy force = weight of displaced water = dx Vxg

where d = density of water V = volume of the ball  ,  g = 10 m/s2deceleration of the body by

buoyancy force, a = (dVg)/ m

where m= d'V            d' = density of block
 
a = dVg/(d' V) = dg/d' =(d/d')*g =g/(0.8)= 10/0.8

(Given, d'/d = 0.8)= 12.5 m/s2

Net deceleration of ball,a' = a-g = 2.5 m/s2Final speed of ball v' = 0  

Use v' = v2 + 2a's      s= depth of ball in the water

=> 40 = 0 + 2x2.5xs    => s = 8m

Q. Equal masses of water and a liquid of relative density 2 are mixed together. Then, the mixture has a relative density of (in g/cm3)  a)2/3 b)4/3 c)3/2 d)3

Ans: The masses of two liquids are equal, let it be m.

Let the relative densities of water and liquid be ρ1 and ρ2 respectively.


The volume of the two be V1 and V2, of water and liquid respectively.


The volume of the mixture would be, V = V1 + V2 (1)


also, volume = mass/density


thus,


2m/ρ (V) = m/ρ1 (V 1 ) + m/ρ2 (V 2 )


here ρ1 = 1,ρ2 = 2 and ρ is the relative density of the mixture.


now, 2/ρ = 1/ρ1 + 1/ρ2


by substituting the values, we ρ/2 = 2/3


or, the relative density of the combined liquid will be, ρ=4/3


More Questions are solved Here:  http://jsuniltutorial.weebly.com/

Sure shoot MCQ for class 9 science term-2 ClicK Here
IX Thrust and Pressure, Archimedes’ Principle, Relative Density
CBSE Class 9 - Science - Chapter 10: Flotation: Notes and Quest
MCQ: Flotation: Thrust, Pressure, Buoyancy and Density
Thrust and Pressure, Archimedes’ Principle, Relative Density key point
Notes : Flotation: Thrust, Pressure, Buoyancy and Density
Physics Flotation Term-II Class IX  Buoyant force Detail Study
9th Physics Solved Numerical Floating bodies

Comments

  1. Physics class
    Chapter Floatation
    The mass of an empty bucket of capacity 10 litres is 1 kg. Find its mass when completely filled with a liquid of relative density 0.7.
    #class9
    Sol.
    V of ticket = 10/1000=0.01m^3
    Density of liquid = RD of liquid x Density of water = 0.7 × 1000kg/m^3 =700kg/m^3
    Mass of liquid = density x volume = 700kg/m^3 × 0.01m^3= 7kg
    Hence, Mass of bucket filled with liquid completely = 7+1=8 kg

    ReplyDelete

Post a Comment

CBSE ADDA :By Jsunil Sir : Your Ultimate Destination for CBSE Exam Preparation and Academic Insights

Class 10 Chapter 02 Acid Bases and Salts NCERT Activity Explanation

NCERT Activity Chapter 02 Acid Bases and Salt Class 10 Chemistry Activity 2.1 Indicator Acid Base Red litmus No Change Blue Blue Litmus Red No change Phenolphthalein Colourless Pink Methyl Orange Pink   Yellow Indictors are substance which change colour in acidic or basic media. Activity 2.2 There are some substances whose odour changes in in acidic or basic media. These are called olfactory indicators. Like onion vanilla, onion and clove. These changes smell in basic solution. Activity 2.3 Take about 5 mL of dilute sulphuric acid in a test tube and add few pieces of zinc granules to it. => You will observe bubbles of hydrogen gas on the surface of zinc granules. Zn + H2SO4 --> ZnSO4 + H2 => Pass the Hydrogen gas through the soap solution. Bubbles formed in the soap solution as Hydrogen gas it does not get dissolved in it

CBSE I NCERT 10th Numerical Problem solved Reflection and reflection of light

Q. 1. A concave mirror of focal length 20cm is placed 50 cm from a wall. How far from the wall an object be placed to form its real image on the wall?  Solution: V= -50 cm F= -20cm From mirror formula 1/u = 1/f – 1/v = -1/20+ 1/50 = - 3/100  U = - 33.3 cm Therefore, the distance of the object from the wall x =  50 – u X = 50 – 33.3 = 16.7 cm. Q.2. An object is placed at a distance of 40cm from a concave mirror of focal length 15cm. If the object is displaced through a distance of 20 cm towards the mirror, By how much distance is the image displaced? Answer: Here f = - 15 cm, u = - 40 cm Now 1/f = 1/u + 1/v Then 1/v = 1/f – 1/u Or V= uf/u-f =( - 40 x -15)/25 = -24 cm Then object is displaced towards the mirror let u1 be the distance object from the Mirror in its new position. Then u1 = -(40-20) = -20cm If the image is formed at a distance u1 from the mirror then v1 = u1f/u1-f = -20X-15/-20+15 = -60 cm. = - 20 x-15/-20+15 = -60 cm. Therefor

Class 10 Metal and Non MetalsChapter 03 NCERT Activity Solutions

X Class 10 NCERT Activity Explanation Class 10 Metals and Non Metals Activity 3.1 Page No. 37 Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample. They have a shining surface. Clean the surface of each sample by rubbing them with sand paper and note their appearance again. They become more shiny. => Freshly cut Metal have shiny surface Activity 3.2 Page No. 37 Take small pieces of iron, copper, aluminium, and magnesium. Try to cut these metals with a sharp knife and note your observations. They are very hard to cut. Hold a piece of sodium metal with a pair of tongs and try to cut it with a knife. Sodium can be cut easily with knife. Hence K and Na are soft metal cut with knife Activity 3.3 Page No. 38 Take pieces of iron, zinc, lead and copper try to strike it four or five times with a hammer. These metals are beaten into thin sheet on hammering. This property of metal is called malleability and metals are called malleable. Activity 3.4 Page

Living science ratna sagar class 6 solutions

Ratna sagar living science 6 answers by jsunil. Class6 Living science solution Term-1 Living Science Solution chapter-1 Source of food Download File Living Science Solution chapter-2 Component of food Download File Living Science Solution chapter-3 Fibre to fabric Download File Living Science Sol ch-4 Sorting of material into group Download File Living Science Soln ch-5 Separation of substance Download File Living Science Solution chapter-6 Change around Us Download File Living Science Solution ch-7 Living and Non Living Download File Living Science Solution ch-8 Getting to Know Plants Download File Living Science Sol ch-9 The Body and Its movements Download File Visit given link for full answer Class6 Living science solution Term-II

Electricity numerical for class 10 CBSE Trend Setter 50 Problems

1. The current passing through a room heater has been halved. What will happen to the heat produced by it? 2. An electric iron of resistance 20 ohm draws a current of 5 amperes. Calculate the heat produced in 30 seconds. 3. An electric heater of resistance 8 ohm takes a current of 15 A from the mains supply line. Calculate the rate at which heat is developed in the heater. 4. A resistance of 40 ohms and one of 60 ohms are arranged in series across 220 volt supply. Find the heat in joules produced by this combination in half a minute. 5. A resistance of 25 ohm is connected to a 12 V battery. Calculate the heat energy in joules generated per minute. 6. 100 joules of heat is produced per second in a 4 ohm resistor. What is the potential difference across the resistor? 7. An electric iron is connected to the mains power supply of 220 V. When the electric iron is adjusted at minimum heating’ it consumes a power of 360 W but at ‘maximum heating’ it takes a power of 840 W. Ca