87576 The rate of decay of certain substance is directly proportional to the amount present at that instant. Initially, there are \(27 \mathrm{gms}\) of certain substance and 3 hours later it is found that \(8 \mathrm{gms}\) are left, then the amount left after one more hour is

87577 The rate at which the metal cools in moving air is proportional to the difference of temperatures between the metal and air. If the air temperature is \(290 \mathrm{~K}\) and the metal temperature drops from \(370 \mathrm{~K}\) to \(330 \mathrm{~K}\) in 10 minutes, then the time required to drop the temperature upto \(295 \mathrm{~K}\) is

87578 the rate of decay of mass of a certain substance at time ' \(t\) ' is proportional to the mass at that instant. The time during which the original mass of \(\mathrm{m}_{0} \mathrm{gm}\). will be left to \(\mathrm{m}_{1} \mathrm{gm}\). is ( \(K\) is constant of proportionality)

87579 If the populations grow at the rate \(5 \%\) per year, then the time taken for the population to become double is (Given \(\log 2=\mathbf{0 . 6 9 1 2}\) )

87576 The rate of decay of certain substance is directly proportional to the amount present at that instant. Initially, there are \(27 \mathrm{gms}\) of certain substance and 3 hours later it is found that \(8 \mathrm{gms}\) are left, then the amount left after one more hour is

87577 The rate at which the metal cools in moving air is proportional to the difference of temperatures between the metal and air. If the air temperature is \(290 \mathrm{~K}\) and the metal temperature drops from \(370 \mathrm{~K}\) to \(330 \mathrm{~K}\) in 10 minutes, then the time required to drop the temperature upto \(295 \mathrm{~K}\) is

87578 the rate of decay of mass of a certain substance at time ' \(t\) ' is proportional to the mass at that instant. The time during which the original mass of \(\mathrm{m}_{0} \mathrm{gm}\). will be left to \(\mathrm{m}_{1} \mathrm{gm}\). is ( \(K\) is constant of proportionality)

87579 If the populations grow at the rate \(5 \%\) per year, then the time taken for the population to become double is (Given \(\log 2=\mathbf{0 . 6 9 1 2}\) )

87576 The rate of decay of certain substance is directly proportional to the amount present at that instant. Initially, there are \(27 \mathrm{gms}\) of certain substance and 3 hours later it is found that \(8 \mathrm{gms}\) are left, then the amount left after one more hour is

87577 The rate at which the metal cools in moving air is proportional to the difference of temperatures between the metal and air. If the air temperature is \(290 \mathrm{~K}\) and the metal temperature drops from \(370 \mathrm{~K}\) to \(330 \mathrm{~K}\) in 10 minutes, then the time required to drop the temperature upto \(295 \mathrm{~K}\) is

87578 the rate of decay of mass of a certain substance at time ' \(t\) ' is proportional to the mass at that instant. The time during which the original mass of \(\mathrm{m}_{0} \mathrm{gm}\). will be left to \(\mathrm{m}_{1} \mathrm{gm}\). is ( \(K\) is constant of proportionality)

87579 If the populations grow at the rate \(5 \%\) per year, then the time taken for the population to become double is (Given \(\log 2=\mathbf{0 . 6 9 1 2}\) )

87576 The rate of decay of certain substance is directly proportional to the amount present at that instant. Initially, there are \(27 \mathrm{gms}\) of certain substance and 3 hours later it is found that \(8 \mathrm{gms}\) are left, then the amount left after one more hour is

87577 The rate at which the metal cools in moving air is proportional to the difference of temperatures between the metal and air. If the air temperature is \(290 \mathrm{~K}\) and the metal temperature drops from \(370 \mathrm{~K}\) to \(330 \mathrm{~K}\) in 10 minutes, then the time required to drop the temperature upto \(295 \mathrm{~K}\) is

87578 the rate of decay of mass of a certain substance at time ' \(t\) ' is proportional to the mass at that instant. The time during which the original mass of \(\mathrm{m}_{0} \mathrm{gm}\). will be left to \(\mathrm{m}_{1} \mathrm{gm}\). is ( \(K\) is constant of proportionality)

87579 If the populations grow at the rate \(5 \%\) per year, then the time taken for the population to become double is (Given \(\log 2=\mathbf{0 . 6 9 1 2}\) )