Home
Class 11
PHYSICS
A lead bullet just melts when stopped by...

A lead bullet just melts when stopped by an obstacle. Assuming that 25 per cent of the heat is absorbed by the obstacle, find the velocity of the bullet if its initial temperature is `27^@C`. (Melting point of lead=`327^@C`, specific heat of lead =`0.03 calories //gm//^@C`, latent heat of fusion of lead=`6calories//gm,J=4.2 joules //calorie`).

Text Solution

Verified by Experts

The correct Answer is:
A, B

75% Heatis retained by bullet
`3/4 [1/2mv^2] = msDeltatheta + mL`
or `v = (sqrt(8sDeltatheta+8L)/3)`
Substituting the values, we have
`v=(sqrt((8 xx 0.03 xx 4.2 xx 300) + (8 xx 6xx 4.2))/3)`
`=12.96 m//s`
Promotional Banner

Topper's Solved these Questions

  • CALORIMETRY & HEAT TRANSFER

    DC PANDEY|Exercise Exercise 22.2|7 Videos
  • CALORIMETRY & HEAT TRANSFER

    DC PANDEY|Exercise Assertion And Reason|10 Videos
  • CALORIMETRY & HEAT TRANSFER

    DC PANDEY|Exercise Miscellaneous Examples|13 Videos
  • BASIC MATHEMATICS

    DC PANDEY|Exercise Exercise|13 Videos
  • CALORIMETRY AND HEAT TRANSFER

    DC PANDEY|Exercise Medical entrance s gallery|38 Videos

Similar Questions

Explore conceptually related problems

A bullet of lead melts when stopped by obstacle. Assuming that 25 per cent of the heat is absorbed by the osbtacle, find the velocity of the bullet if its initial temperature is 27^(@)C . (Melting point of lead 327^(@)C , specific heat capacity of lead =30 cal kg^(-1)K^(-1) specific latent heat of fusion of lead =6000 cal kg^(-1) and J=4.2 joules cal^(-1) )

During the Indo-Pakistan war, a soldier observed that the lead bullet fired by him just melted, when stopped by an obstacle. Assuming that whole of the kinetic energy of bullet got converted into heat energy, calculate the velocity of the bullet. Given that initial temperature of the bullet = 47.6^(@)C . Melting point of lead 321^(@)C , specific heat of lead = 0.3 cal g^(-1).^(@)C^(-1) , latent heat of lead = 6 cal g^(-1) and J = 4.2 xx 10^(7) erg cal^(-1) .

A 25 g bullet travelling at 400 m//s passes through a thin iron wall and emerges at a speed of 250 m/s. If the bullet absorbs 50 percent of the heat generated, a. what will be the temperature rise of the bullet? B. If the ambient temperature is 20^(@)C will the bullet melt, and if so how much? Given that specific heat of head is 0.03 "cal"//gm^(@)C and latent heat of fusion of lead is 6000 cal/kg. It is also given that melting pont of lead is 320^(@)C .

A lead bullet penetrates into a solid object and melts Assuming that 50% of its K.E. was used to heat it , calculate the initial speed of the bullet , The initial temp , of bullet is 27^(@)c and its melting point is 327^(@)C Latent heat of fasion of lead = 2.5 xx 10^(4) J kg^(-1) and sp heat capacity of lead = 1.25 J kg^(-1) K^(-1)

A lead bullet of 10g travelling at 300m//s strikes against a block of wood and comes to rest. Assuming 50% heat is absorbed by the bullet, the increase in its temperature is (sp-heat of lead is 150J//Kg-K )

If a lead bullet be suddenly stopoed and all its energy be used to heat it, with what velocity must the bullet be fired to raise the temperature through 100^(@)C ? (Specific heat capacity of leat =31.4 calories per kg per kelvin and J=4.2 joules per calorie.)

A 2 g bullet moving with a velocity of 200 m/s is brought to a sudden stoppage by an obstacle. The total heat produced goes to the bullet. If the specific heat of the bullet is 0.03 cal//g^@C , the rise in its temperature will be

A lead bullet strikes a target with a velocity of 500 ms^(-1) and the bullet falls dead. Calculate the rise in temperature of the bullet assuming that 40% of the heat produced is used in heating the bullet. Given specific heat of lead = 0.03 cal g^(-1).^(@)C^(-1), J = 4.2 J cal^(-1) .

Four cubes of ice at -10^(@)C each one gm is taken out from the refrigerator and are put in 150 gm of water at 20^(@)C . The temperature of water when thermal equilibrium is attained. Assuming that no heat is lost to the outside and water equivalent of contaner is 46 gm . (Specific heat capacity of water = 1 cal//gm-^(@)C , Specific heat capacity of ice =0.5 cal//gm-^(@)C , Latent heat of fusion of ice = 80 cal//gm-^(@)C )