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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
A beaker contains 200 g of water. The heat capacity of the beaker is equal to that of 20 g of water. The initial temperature of water in the beaker is `20^@C` .If 440 g of hot water at `92^@C` is poured in it, the final temperature (neglecting radiation loss) will be nearest toA. `58^@ C`B. `68^@ C`C. `73^@ C`D. `78^@ C` |
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Answer» Correct Answer - B From principle of calorimetry `m_("water") xx S_(w) xx (theta-20)+(mS)xx(theta-20)` =`m_("hot water") xx S_(w) xx(92 -theta)`. |
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| 102. |
A copper ring has a diameter of exactly `25 mm` at its temperature is `0^@ C`. An aluminium sphere has a diameter is exactly `25.05 mm` and its temperature it `100^@ C`. The sphere is placed on top of the rind and two allowed to come to thermal equilibrium. The ratio of the mass of the sphere and ring is (given `alpha_(cu) = 17 xx 10^(-6).^@C^(-1)alpha_(Al) = 2.3 xx 10^(-5) .^@C^(-1)` specific heat of `Cu` is `0.0923 cal//g^@C` and for `Al` is `0.215 cal//g^@C`.A. `1//5`B. `23//108`C. `23//54`D. `216//23` |
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Answer» Correct Answer - C A equilibrium `(MsDelta t)_(Cu) = (MsDelta t)_(Al)` When sphere passing through ring then Radius of sphere = Radius ring `(l_(2))_(Cu) = (l_(2))_(Al)` `25(1+alpha_(Cu) Deltat)=25.05(1+ alpha_(Al) Delta t)`. |
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| 103. |
A colirimeter of water equiualent `15g` containe `165g` of water at `25^(@)C sleam at `100^(@)C `is passed through the water for same time .The temperature is increases to 30^(@)C` and the mass orf the calorimater and its contents is increased by `1.5g` calculate the specfic latent head of toporition of water Specific head calacity of water is `1cal g ^(-1)^(@)C^(-1)` |
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Answer» Let L be the specific intent head of vaporization of water The mass of the steam condensed is `1.5g` Head lost in condensation of steam is `Q = (1.5g)L` The condensed water cool from `100^(@)C to 30^(@)C `head lose in this process is m `Q = (1.5g)(1 calg^(-1)^(@)C)= 105 cal` Head supplies to the calorimater and to the cold water the srise in temperature from `25^(@)Cto `Q_(s) = (1.5g + 165 g)(1calg^(-1)^(@)C ^(-1))(5^(@)C) = 900 cal` If no head is lost to the surerounding `(1.5g)L + 105 cal= 900 cal` `or L = 530 cal g` |
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| 104. |
A uniform brass disc of radius a and mass m is set into spinning with angular speed `omega_0` about an axis passing through centre of disc and perpendicular to the palne of disc. If its temperature increases from `theta_1^@C` to `theta_2^@C` with out disturbing the disc, what will be its new angular speed ?A. `omega_0[1+2a(theta_2-theta_1)]`B. `omega_0[1+a(theta_2-theta_1)]`C. `(omega_0)/([1+2alpha(theta_2-theta_2)])`D. none of these |
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Answer» Correct Answer - C From the conservation priciple of angular momentum `L_i=L_f` `I_0omega_0=Iomega` Here, `I=I_0[1+2alpha(theta_2-theta_1)]` `omega=(I_0omega_0)/(I_0[1+2alpha(theta_2-theta_1)])=(omega_0)/([1+2alpha(theta_2-theta_1)])` |
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| 105. |
On a winter day temperature of the tap water is `20^(@)C` where as the room temperature is `5^(@)C` water is stored in a tank of capacity `0.5 m^(3)` for household use .If it were possible to use the beat liberated by the water to lift a `10kg` mass vertically , how high can it be lifted as the water comes to the room temperature ? Take `g = 10ms^(-2)` |
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Answer» Correct Answer - A::C Here `m == 0.5m^(2) = 500L = 500kg` So the heat liberated during the water changes`20^(@)C to 5^(@)C` `= 500 xx 4200 xx 15`[Delta theta = 20- 5= 15]` `= 500 xx 4200 xx 15` `= 75 xx 420 xx 1000` `= 31500 xx 1000` Let the height = h` The required work `= mgh = 10 xx 10 xx h= 100h` But,`100h = 315000m` `rArr h = 31500 m = 315km` |
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| 106. |
A calorimeter contains 0.2 kg of water at `30^@C`. 0.1 kg of water at `60^@C` is added to it, the mixture is well stirred and the resulting temperature if found to be `35^@C`. The thermal capacity of the calorimeter isA. `6300 J//K`B. `1260 J//K`C. `4200 J//K`D. none of these |
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Answer» Correct Answer - B Let X be The thermal capacity of calorimeter and specific heat of water `=4200 J//kg-K` Heat lost by 0.1 kg of water = Heat gained by water in calorimeter+ Heat gained by calorimeter `implies0.1xx4200xx(60-35)` `=0.2xx4200xx(35-30)+X(35-30)` `10500=4200+5 X implies X=1260 J//K` |
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| 107. |
The temperature of a silver bar rised by `10.0^@`C when it absorbs `1.23kJ` of energy by heat. The mass of bar is 525g. Determine the specific heat of silver. |
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Answer» We find its specific heat from the definition, which is contained in the equation `Q=mc_(silver)DeltaT` For energy input by heat to produce a temperature change. Solving we have `c_(silver)=(Q)/(mDeltaT)` `c_(silver)=(1.23xx10^3J)/((0.525kg)(10.0C))=234J//kg.^@C` |
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| 108. |
A 60 kg boy running at `5.0 m//s` while playing basketball falls down on the floor and skids along on his leg until he stopes. How many calories of heat are generated between his leg and the floor? Assume that all this heat energy is confined to a volume of `2.0cm^3` of his flesh. What will be temperature change of the flesh? Assume `c=1.0(cal)/(g^@C)` and `rho=950(kg)/(m^3)` for flesh. |
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Answer» Kinetic energy of boy `K=(1)/(2)mv^2=(1)/(2)(60)(5)^2=750J` Changing this energy in joule into calorie `Q_1=(K)/(J)=(750)/(4.2)=179J` .(i) As this kinetic energy changes into heat energy `(Q_2)` and this heat is confined to flesh of boy, `Q_2=mcDeltaT=(rhoV)cDeltaT` .(ii) As `Q_1=Q_2`, hence from Eqs. (i) and (ii) `179=0.95xx2xx2xxDeltaTimpliesDeltaT=94^@C` |
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| 109. |
A `50 kg` man is running at a speed of `18 kmH^(-1)` If all the kinetic energy of the man be uses to increase the temperature of water from `30^(@)C`how much water can be beated with this energy? |
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Answer» Correct Answer - A K.E. of the mass `== (1)/(2) mV^(2)` `= ((1)/(2)) 50 xx 5^(2)` `= 25 xx 25 625J` The amount of heat required to raise the temperature of water from `20^(@)C to 30^(@)C ` `msDelta theta= m xx(30 - 20)` `4200m` `But, `42 xx 10^(3) m = 625` `rArr m = (625)/(42) xx 10^(-3)` ` = 14.88 xx 10^(-3) kg` `15 g` |
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| 110. |
The variation of length of two metal rods A and B with change in temperature is shown in Fig. the coefficient of linear expansion `alpha_A` for the metal A and the temperature T will beA. `alpha_(A)=3 xx 10^(-6)//^(@)C,500^(@)C`B. `alpha_(A)=3 xx 10^(-6)//^(@)C,222.22^(@)C`C. `alpha_(A)=27 xx 10^(-6)//^(@)C,500^(@)C`D. `alpha_(A)=27xx 10^(-6)//^(@)C,222.22^(@)C` |
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Answer» Correct Answer - D Slope of line `A=((1006-1000)mm)/(T^@C)=(DeltaL)/(DeltaT)=Lalpha_A` i.e., `(6)/(T)mm//^(@)C=(1000mm)alpha_A` similary for line B `(2)/(T)mm//^(@)C=(1002mm)alpha_B` Dividing Eq. (i) by Eq. (ii) `3=(1000 alpha_A)/(1002alpha_B)approx=3alpha=3alpha_B` From Eq. (iii) `alpha_A=3xx9xx10^-6=27xx10^(-6)//^(@)C` from eq. (i) `T=(6)/(1000alpha_A)=(6xx10^6)/(1000xx27)` `=222.22^@C` |
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| 111. |
A vessel contains M grams of water at a certain temperature and water at certain other temperature is passed into it at a constant rate of `mg//s`. The variation of temeprature of the mixture with time is shown in Fig. The values of M and m are, respectively (the heat exhanged after a long time is 800 cal)A. 40 and 2B. 40 and 4C. 20 and 4D. 20 and 2 |
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Answer» Correct Answer - B Evidently the initial temperature of the water contained in the vessel (Mg) is `80^@C` and the temperature of the water passed into it is `60^@C` as the final temperature of the mixture tends to attain a value of `60^@C` `Mxx1(80-70)=mxx10xx1(70-60)` or `M//m=10` since the heat exhanged after a long time is 800 cal. `(Mg)(1cal//(m^(@)C)(80-60^(@)C)=80cal` or `M=40g` `implies m=4g` |
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| 112. |
Two plates identical in size, one of black and rough surface `(B_1)` and the other smooth and polished `(A_2)` are in terconnected by a thin horizontal pipe with a mercury pellet at the centre. Two more plates `A_1` (identical to `A_2`) and `B_2` (identical `B_1`) are heated to the same temperature and placed closed to the plates `B_1`, and `A_2` shown in Fig. The mercury pelletA. moves to the rightB. moves to the leftC. remains stationaryD. starts oscillating left and right |
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Answer» Correct Answer - C Smooth and polished plates are poor radiators of heat. Hence, heat coming out from A is small, even though B being a black and rough plate is a good absorber. Effectively the heat coming to the left of pellet P is small. Black and rough plates are good radiators of heat. Hence plate `B_2` radiates heat to a satisfactory level , however plate `A_2` being smooth and polisher, is a bad absorber.Effectively the heat coming to the right of P is also small. |
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| 113. |
Rishi is surprised when he sees water boiling at 115°C in a container. Give reasons as to why water can boil at the above temperature. |
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Answer» The water boils at the higher temperature because of the reasons given below : 1. The water used by Rishi might be impure. The boiling of a liquid increases with the addition of impurities. 2. Rishi might have used a container which creates a pressure within. The boiling point of a liquid increases with an increase in pressure. |
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| 114. |
A aluminium can of cylindrical shape contains `500cm^3` of water. The area of the inner cross section of the can is `125cm^2`. All measurements refer ti `10^@C`. Find the rise in the water level if the temperature increases to`80^@C`. The coefficient of linear expansion of aluminium `=23 xx 10^(-6 @ )C(-1)` respectively. |
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Answer» It is given that at `10^@C` volume of beer is `500cm^3` and the area of cross section of can is `125cm^@`. Thus height of bear level is `h=(500)/(125)=4cm` Now at `80^@C`, volume of beer becomes `V_(80^@C)=500(1+3.2xx10^-4xx70)` `=511.2cm^3` At `80^@C` area of cross section of can becomes At `80^@C=125[1+2alpha_(A1)xx70]` `=125[1+2xx2.3xx10^-5xx70]` `=125.402cm^2` Thus new height of beer level at `80^@C` is `h^(`)=(V_(80^@C))/(A_(80^@C))=(511.2)/(125.402)=4.076cm` Thus rise in level of beer is `Deltah=h^(`)-h=4.076-4.0=0.076cm` |
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| 115. |
A rectangular block is heated from `0^(@)C` to `100^(@)C`. The percentage increases in its length is 0.10 % what will be the percentage increases in it volume ?A. `0.03%`B. `0.10%`C. `0.30%`D. none of these |
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Answer» Correct Answer - C % increases in volume =3x0.1=0.3% |
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| 116. |
A thin copper wire of length l increases in length by 1% when heated from `0^(@)C` to `100^(@)C`. If a then copper plate of area `2lxxl` is heated from `0^(@)C` to `100^(@)C`, the percentage increases in its area will beA. `1%`B. `2%`C. `3%`D. `4%` |
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Answer» Correct Answer - B % increases in area `(DeltaA)/Axx100=B xx Delta thetaxx100` `=2xx(axx Delta theta)xx100=2xx1=2%` |
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| 117. |
Ice at `0^@C` is added to 200 g of water initially at `70^@C` in a vacuum flask. When 50 g of ice has been added and has all melted the temperature of the flask and contents is `40^@C`. When a further 80 g of ice has been added and has all melted the temperature of the whole becomes `10^@C`. Find the latent heat of fusion of ice.A. `80 cal//g`B. `90 cal//g`C. `70 cal//g`D. `540 cal//g` |
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Answer» Correct Answer - B According to principle of calorimetry `ML_f+MsDeltaT=(msDeltaT)_(water)+(msDeltaT)_(flask)` `50L_F+50xx1xx(40-0)` `200xx1xx(70-40)+W(70-40)` `50L_F+2000=(200+W)30` `5L_F=400+3W` .(i) Now the system contains `(200+50)` g of water at `40^@C`, so when further 80 g of ice is added `80L_F+80xx1xx(10-0)` `=250xx1xx(40-10)+W(40-10)` `80L_F=670+3W` .(ii) solving Eqs. (i) and (ii), `L_F=90(cal)/(g)` and `W=(50)/(3)g` |
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| 118. |
What impact will climate changes have on the crops of food? |
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Answer» At the present rate of increase of green house effect, it is expected that nearly 30% of the plant species will extinct by the year 2050 and up to 70% by the end of the year 2100. In the near future, warming of nearly 3oC will result in poor yield in farms in low latitude regions. This will increase the rise of malnutrition. |
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| 119. |
How will climate changes affect the various animal species? |
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Answer» At the present rate of increase of green house effect, it is expected that nearly 30% of animal species will extinct by the year 2050 and up to 70% by the end of the year 2100. This will disturb ecosystem. The animals from the equatorial region will shift to higher latitude in search of ice and cold region. The absorption of carbon dioxide by the ocean will cause acidification due to which marine species will migrate. |
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| 120. |
Why does the weather become very cold after a hail storm? Explain. |
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Answer» Reasons used : 1kg of ice on meting absorbs 336000J of heat energy and 1kg of water to freeze will absorb 336000J of heat energy. After hail storm, to melt ice balls very large amount of heat is extracted from surroundings (sp. heat capacity of ice 336000J) hence temp, falls and it becomes very cold. |
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| 121. |
The amount of heat energy required to melt a given mass of a substance at its melting point without any rise in temperature is called : (a) heat capacity (b) sp. heat capacity (c) latent heat of fusion (d) sp. latent heat of fusion |
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Answer» (c) latent heat of fusion |
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| 122. |
What do you understand by the term greenhouse effect? |
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Answer» Green house effect : Sun rays from the Sun pass the earth’s atmosphere and infrared radiations of short wave length reach the earth’s surface and objects (plants) on it. They get warmed during day time. At night the same earth’s At mosphere becomes opaque i.e. does not allow infra-red radiations of long wavelength to go back. In other words atmosphere entraps (or long wavelengths are absorbed by green house gases like CO2 methane, chlorofluorocarbons) and hence atmosphere acts as green house with glass walls and raises the temp, inside. Hence green house effect “is the phenomenon in which infrared radiations of long wavelength given out from the surface of earth are absorbed by its atmospheric gases to keep the environment at the earth’s surface and its lower atmosphere warm”. |
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| 123. |
Name the two main greenhouse gases and how they enter the atmosphere. |
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Answer» Two main green house gases are : 1. Carbon dioxide CO2 2. Methane gas CH4 CO2 enters the atmosphere through 1. Fossil fuel based power plants 2. Deforestation 3. Internal combustion engines. 4. Increasing population and their activities. Methane (CH4) enters the atmosphere when dead vegetable matter decays. It is mainly produced due to the decaying dead plant remains in the paddy fields. It is also produced in marshly lands, sewage, coal mines and bio gas plants. |
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| 124. |
Is the specific heat capacity of ice greater, equal to or less than water? |
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Answer» Specific heat capacity of ice is less than specific heat capacity of water. |
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| 125. |
A heat flux of 4000 J/s is to be passed through a copper rod of length 10 cm and area of cross section `100cm^2`. The thermal conductivity of copper is `400W//m//^(@)C` The two ends of this rod must be kept at a temperature difference ofA. `1^@C`B. `10^@C`C. `100^@C`D. `1000^@C` |
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Answer» Correct Answer - C From `(dQ)/(dt)=(KaDeltatheta)/(l)` `implies Delta theta=(l)/(KxxA)xx(dQ)/(dt)=(0.1)/(400xx(100xx10^-4))xx4000=100^@C` |
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| 126. |
Differentiate between heat and temperature. |
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| 127. |
Define temperature and name its S.I. unit. |
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Answer» The quantity which determines the direction of flow of heat between two bodies kept in contact is called temperature. S.I. unit kelvin (K). |
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| 128. |
In an experiment, 17g of ice is used to bring down the temperature of 40 g of water at 34℃ to its freezing temperature. The specific heat capacity of water is 4.2 J kg-1 K-1. Calculate the specific latent heat of ice. State one important assumption made in the above calculation. |
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Answer» Mass of ice m1 = 17 g Mass of water m2 = 40 g. Change in temperature = 34 – 0 = 34K Specific heat capacity of water is 4.2Jg-1K-1. Assuming there is no loss of heat, heat energy gained by ice (latent heat of ice), Q= heat energy released by water Q = 40 × 34 × 4.2 = 5712 J. Specific latent heat of ice = L = Q/m = 5712/17 = 336Jg-1 |
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| 129. |
A van of mass `1500kg` travelling at a speed of `54 km h^(-1)` is stopped in `10 s` Assuming that all the machantical energy lost appeats as thermal energy in the brake mechanical find the evarage in the brake of thermal energy in cal `s^(-1)` |
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Answer» Correct Answer - A::C::D Mass of van `= 1500kg` `speed V = 54km//h` `= 54 xx ((5)/(18)) = 15 m//s` `Total K.E. = (1)/(2) mv^(2)` ` = (1)/(2)xx 1500xx (15)^(2)` ` = 750xx 225` = 168750J` `= 40178cal` Loss in the tatal energy `= 40175 cal` In `1` second loss in the total energy `= (40178)/(10) = 4017.8` ` = 4000 cal//sec` THe avarage thermal energy cal//sec `= 4000 cal//sec |
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| 130. |
The tempetature of a solidobject is observed to be constant during a period .In this periodA. heat may have supplied to the bodyB. heat may have been extracted from the bodyC. no heat is supplied to the bodyD. no heat is extracted from the body |
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Answer» Correct Answer - A::B heat may have supplied to the body ,heat may have been extracted from the body |
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| 131. |
The temperature of an object is observed to rise in a period. In this period (a) heat is certainly supplied to it (b) heat is certainly not supplied to it (c) heat may have been supplied to it (d) work may have been done on it. |
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Answer» The correct answer is (c) (d) (c) heat may have been supplied to it (d) work may have been done on it. |
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| 132. |
The temperature of an object is observed to rise in a period. In this period (i) Heat is certainly supplied to it (ii) Heat is certainly not supplied to it (iii) heat may have been supplied to it (iv) work may have been done on it.A. heat is certainly supplied to itB. heat is certainly not supplied to itC. heat may have been supplied to itD. work may have been done on it |
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Answer» Correct Answer - C::D heat may have been supplied to it,work may have been done on it |
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| 133. |
Heat and work are equivalent. This means, (a) when we supply heat to a body we do work on it (b) when we do work on a body we supply heat to it (c) the temperature of a body can be increased by doing work on it (d) a body kept at rest may be set into motion along a line by supplying heat to it. |
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Answer» (c) the temperature of a body can be increased by doing work on it |
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| 134. |
Complete the following sentences:(a) When ice melts, its volume _________(b) Decrease in pressure over ice ________ its melting point.(c) Increase in pressure _________ the boiling point of water.(d)A pressure cooker is based on the principle that boiling point of water increases with the _________(e) The boiling point of water is defined as _________ |
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Answer» (a) When ice melts, its volume decreases. (b) Decrease in pressure over ice increases its melting point. (c) Increase in pressure increases the boiling point of water. (d)A pressure cooker is based on the principle that boiling point of water increases with the increase in pressure. (e) The boiling point of water is defined as the constant temperature at which water changes to steam. |
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| 135. |
An aluminium vessel of mass `0.5kg` contains `0.2kg` of water at `20^(@)C` A block of iron of mass `0.2kg at 100^(@)C` is gently put into the water .Find the equilibrium temperature of the mixture,Specific beat capactities of aluminium , iron and water are `910 J kg^(-1)K^(-1) 470J kg^(-1)K^(-1) and 420J kg^(-1)K^(-1)` respectively |
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Answer» Correct Answer - C Given mass of aluminium `= 0.5 kg` Mass of water` = 0.2 kg` Mass of iron` = 0.2 kg` Tempetature of aluminium and water `= 20^(@) = 293^(@) K` Tempetature of iron `= 100^(@)C = 373 k` Specfic beat of AI ` = 910 J// kg-K` Heat gain ` = 0.5 xx 610 (T - 293) + 0.2 xx 4200 xx (T - 293)` `= (T- 293)(0.5 xx 910 + 0.2 xx 4200)` Heat lost `= 0.2xx 470 xx (373 - T)` We know heat gain = Head lost `rArr (T - 293)(0.5 xx 910 + 0.2 xx 4200)` `= 0.2 xx 470 xx (373 - T)` `rArr(T - 293)(455 + 840)= 94(373 - T)` `rArr (T - 293) (1295)/(94) = (273 - T) [(1295)/(94) =- 14]` ` (T - 293) xx 14 = 373 - T` `rArr 14T - 293 xx 14 = 373 - T` rArr 15T = 373 + 4102 = 4475` `rArr T = (4475)/(15) = 298 k ` ` :. T = (298 - 273)^(@)C = 25^(@)C` `:. ` The final temp ` = 25^(@)C` |
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| 136. |
Water equivalent of a body is measured in (a) kg (b) calorie (c) kelvin (d) m3. |
| Answer» The correct answer is (a) kg | |
| 137. |
Head and work are equivalent.This meansA. when we supply heat to a body we do work on itB. when we do work on a body we supply heat to itC. the temperature of a body can be increased by doing work on itD. a body at rest may be can be set into motion along a line by suppliying heat to it |
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Answer» Correct Answer - C the temperature of a body can be increased by doing work on it |
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| 138. |
The specific heat capacity of a body depends on(a) the heat given (b) the temperature raised(c) the mass of the body (d) the material of the body. |
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Answer» (d) the material of the body. |
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| 139. |
What do you understand by the term latent heat? |
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Answer» Latent heat: The heat energy exchanged in change of phase is not externally manifested by any rise or fall in temperature, it is considered to be hidden in the substance and is called the latent heat. |
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| 140. |
The mechanical equivalent of headA. has the same dimension as heatB. has the same dimension as workC. has the same dimension as energyD. is dimensionless |
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Answer» Correct Answer - D is dimensionless |
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| 141. |
State the effect of increase of pressure on the melting point of ice. |
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Answer» The melting point of ice decreases by the increase in pressure. The melting point of ice decrease by 0.0072oC for every one atmosphere rise in pressure. |
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| 142. |
Which of the following pairs represent units of the same physical quantity ? (a) kelvin and joule (b) kelvin and calorie (c) newton and calorie (d) joule and calorie. |
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Answer» (d) joule and calorie. |
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| 143. |
Name a liquid which has the highest specific heat capacity. |
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Answer» Water has the highest specific heat capacity. |
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| 144. |
Write the approximate value of specific heat capacity of water in S.I. unit. |
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Answer» Specific heat capacity of water = 4200 J kg-1K-1. |
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| 145. |
Write the approximate value of specific latent heat of ice. |
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Answer» Specific latent heat of ice: 336000 J kg-1. |
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| 146. |
What is the principle of the method of mixtures?, |
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Answer» When there is no loss or gain of heat from surroundings, heat lost by hot body or bodies is equal to heat gained by cold body or bodies. |
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| 147. |
Name the law on which principle of methods of mixtures is based. |
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Answer» The principle of mixture is based on the law of conservation of energy. |
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| 148. |
Which of the following pairs of physical quantities may be represented in the same unit ? (a) heat and temperature (b) temperature and mole (c) heat and work (d) specific heat and heat. |
| Answer» (c) heat and work | |
| 149. |
State the effect of an increase of impurities on the melting point of ice. |
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Answer» Increasing the impurities causes the melting point of ice to decrease. |
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| 150. |
Explain how is land breeze caused? |
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Answer» Land breeze : Blowing of cold air from land towards sea. During night temp, of land falls more rapidly as compared to water. Since water has high sp. heat capacity Pressure over sea water decrease and hence air blows from land (high pressure) towards sea (low pressure). |
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