Just what will end up being the payment error when it level can be used (a) on the a summertime time when the climate is cuatro6°C and you will (b) into the a cold weather time in the event that temperature is six°C? Coefficient away from linear extension off material = eleven ? 10 –six °C –step one .

Answer:

(a) Let the correct length measured by a metre scale made up of steel 16 °C be L. Initial temperature, t₁ = 16 °C Temperature on a hot summer day, t₂ = 46 °C aˆ‹So, change in temperature, ?? = t₂

?= step one.1 ? 10 –5 °C aˆ‹-step 1 Therefore, improvement in length, ?L = L ??? = L ? step one.step one ? 10 –5 ? 30

% of error =?LL?100% =L? ??L?100% =1.1?10-5?30?100% =3.3?10-2%(b) Temperature on a winter day, t₂ = 6 °Caˆ‹ aˆ‹So, change in temperature, ?? = t₁

Matter sixteen:

A good metre scale made from material checks out truthfully at the 20°C. Inside a painful and sensitive check out, ranges perfect as much as 0.055 mm in step 1 yards are needed. Discover list of temperature where the try out are performed using this type of metre scale. Coefficient off linear extension away from metal = eleven ? ten –six °C –step one .

Answer:

Given: Temperature at which a metre scale gives an accurate reading, T₁ = 20 °C The value of variation admissible, datehookup app ?L = 0.055 mm = 0.055 ? 10 –3 m, in the length, L₀ = 1 m Coefficient of linear expansion of steel, ? = 11 ? 10 –6 °C –1 Let the range of temperature in which the experiment can be performed be T₂. We know: ?L = L₀ ??T ?0.055?10-3=1?11?10-6?T1±T2?5?10-3=20±T2?10-3?20 ± T2=5?Either T2=20+5=25 °C or T2=20-5=15 °CHence, the experiment can be performed in the temperature range of 15 °C to 25 °C .

Question 17:

The fresh new thickness from liquid on 0°C try 0.998 grams cm –3 as well as cuatro°C is actually step 1.one hundred thousand g cm –step one . Determine an average coefficient out of regularity expansion out-of drinking water from the temperature a number of 0 to 4°C.

Answer:

Given: Density of water at 0°C, ( f₀)= 0.998 g cm -3 Density of water at 4°C, aˆ‹(f₄) = 1.000 g cm-3 Change in temperature, (?t) = 4 o C Let the average coefficient of volume expansion of water in the temperature range of 0 to 4°C be ?.

?=-5?10-cuatro oC-step one For this reason,the typical coefficient from volume expansion off h2o throughout the temperature listing of 0 to help you cuatro°C might possibly be

Matter 18:

Discover ratio of lengths off an iron pole and you may an aluminum rod in which the difference in the lengths was independent out of heat. Coefficients off linear expansion regarding iron and you will aluminum was a dozen ? 10 –6 °C –step 1 and you may 23 ? 10 –six °C –step 1 correspondingly.

Answer:

Let the original length of iron rod be L _Fe and L ‘ aˆ‹ aˆ‹aˆ‹ _Fe be its length when temperature is increased by ?T. Let the original length of aluminium rod be L _Al and L ‘ aˆ‹ aˆ‹aˆ‹ _Al be its length when temperature is increased by ?T.

L’Fe=LFe step one+?Fe??Tand L’Al=LAl 1+?Al??T?L’Fe-L’Al=LFe-LAl+LFe??Fe?T-LAl??Al??T-(1)Given:L’Fe-L’Al=LFe-LAlHence, LFe?Fe=LAl ?Al [using (1)]?LFeLAl=2312The proportion of lengths of your own metal for the aluminum rod is .

Question 19:

An excellent pendulum clock reveals right time within 20°C in the a place in which g = 9.800 m s –2 . The fresh new pendulum consists of a white material pole linked to an effective big ball. It is delivered to another place where grams = nine.788 meters s –1 . From the what heat will the clock inform you right time? Coefficient out-of linear expansion from metal = 12 ? ten –6 °C –step 1 .