143115
What is the number of water molecules present in a tiny drop of water (volume $0.0018 \mathrm{ml}$ ) at room temperature?
1 $4.84 \times 10^{17}$
2 $4.184 \times 10^{18}$
3 $6.023 \times 10^{19}$
4 $6.023 \times 10^{23}$
Explanation:
C Given data, Volume of water drop $=0.0018 \mathrm{~m} l$ Density of drop $=1000 \mathrm{~kg} / \mathrm{m}^{3}=1 \mathrm{~g} / \mathrm{m} l$ Mass of $0.0018 \mathrm{~m} l$ water $=$ Density $\times$ Volume $=1 \times 0.0018$ $=0.0018 \mathrm{~g}$ $\because$ We know that, Molar mass of water, $\mathrm{M}_{\mathrm{H}_{2} \mathrm{O}} =(2 \times 1)+16$ $=2+16$ $=18 \mathrm{~g} / \mathrm{mol}$ No. of moles is drop of water, $=\frac{\text { Mass }}{\text { Molar mass }}=\frac{0.0018}{18}$ $=0.0001 \mathrm{~mol}$ $\therefore$ No. of molecule present in drop of water $=6.023 \times 10^{23} \times 0.0001$ $\quad\left\{\because \mathrm{N}_{\mathrm{A}}=6.023 \times 10^{23} \text { molecule in } 1 \text { mole }\right\}$ $=6.023 \times 10^{19}$
NDA (II) 2008
Mechanical Properties of Fluids
143073
The radius of a soap bubble is increased from 1 $\mathrm{cm}$ to $4 \mathrm{~cm}$. If the surface tension of the soap solution is $0.021 \mathrm{Nm}^{-1}$, then the work done to increase the radius is
143077
If two soap bubbles of different radii are connected by a tube, then
1 Air flows from bigger bubble to the smaller bubble till sizes become equal
2 Air flows from bigger bubble to the smaller bubble till sizes are interchanged
3 Air flows from smaller bubble to bigger
4 There is no flow of air
Explanation:
C The additional pressure inside a soap bubble is inversely proportional to the radius of the soap bubble. $\mathrm{P} \propto \frac{1}{\mathrm{r}}$ $\mathrm{r}$ being the radius of bubble. If the pressure inside a smaller bubble is greater than the pressure inside a larger bubble. Thus if these two bubbles are connected by a tube, air will flow from the smaller bubble to the larger bubble and the larger bubble arouse at the Expense of the smaller bubble.
BCECE-2004
Mechanical Properties of Fluids
143019
A lead shot of $1 \mathrm{~mm}$ diameter falls through a long column of glycerine. The variation of its velocity $v$ with distance covered is represented by :
1
2
3 covered
4
Explanation:
A When lead shot of $1 \mathrm{~mm}$ diameter falls in glycerine through a long column, then due to gravity velocity of lead shot initially increases for some time. After some time due to viscosity of glycerine lead shot will attain a constant terminal velocity. Since, some upthrust force acting on lead shot due to glycerine so velocity will not be fully linear. So, the variation in velocity is given in figure.
143115
What is the number of water molecules present in a tiny drop of water (volume $0.0018 \mathrm{ml}$ ) at room temperature?
1 $4.84 \times 10^{17}$
2 $4.184 \times 10^{18}$
3 $6.023 \times 10^{19}$
4 $6.023 \times 10^{23}$
Explanation:
C Given data, Volume of water drop $=0.0018 \mathrm{~m} l$ Density of drop $=1000 \mathrm{~kg} / \mathrm{m}^{3}=1 \mathrm{~g} / \mathrm{m} l$ Mass of $0.0018 \mathrm{~m} l$ water $=$ Density $\times$ Volume $=1 \times 0.0018$ $=0.0018 \mathrm{~g}$ $\because$ We know that, Molar mass of water, $\mathrm{M}_{\mathrm{H}_{2} \mathrm{O}} =(2 \times 1)+16$ $=2+16$ $=18 \mathrm{~g} / \mathrm{mol}$ No. of moles is drop of water, $=\frac{\text { Mass }}{\text { Molar mass }}=\frac{0.0018}{18}$ $=0.0001 \mathrm{~mol}$ $\therefore$ No. of molecule present in drop of water $=6.023 \times 10^{23} \times 0.0001$ $\quad\left\{\because \mathrm{N}_{\mathrm{A}}=6.023 \times 10^{23} \text { molecule in } 1 \text { mole }\right\}$ $=6.023 \times 10^{19}$
NDA (II) 2008
Mechanical Properties of Fluids
143073
The radius of a soap bubble is increased from 1 $\mathrm{cm}$ to $4 \mathrm{~cm}$. If the surface tension of the soap solution is $0.021 \mathrm{Nm}^{-1}$, then the work done to increase the radius is
143077
If two soap bubbles of different radii are connected by a tube, then
1 Air flows from bigger bubble to the smaller bubble till sizes become equal
2 Air flows from bigger bubble to the smaller bubble till sizes are interchanged
3 Air flows from smaller bubble to bigger
4 There is no flow of air
Explanation:
C The additional pressure inside a soap bubble is inversely proportional to the radius of the soap bubble. $\mathrm{P} \propto \frac{1}{\mathrm{r}}$ $\mathrm{r}$ being the radius of bubble. If the pressure inside a smaller bubble is greater than the pressure inside a larger bubble. Thus if these two bubbles are connected by a tube, air will flow from the smaller bubble to the larger bubble and the larger bubble arouse at the Expense of the smaller bubble.
BCECE-2004
Mechanical Properties of Fluids
143019
A lead shot of $1 \mathrm{~mm}$ diameter falls through a long column of glycerine. The variation of its velocity $v$ with distance covered is represented by :
1
2
3 covered
4
Explanation:
A When lead shot of $1 \mathrm{~mm}$ diameter falls in glycerine through a long column, then due to gravity velocity of lead shot initially increases for some time. After some time due to viscosity of glycerine lead shot will attain a constant terminal velocity. Since, some upthrust force acting on lead shot due to glycerine so velocity will not be fully linear. So, the variation in velocity is given in figure.
143115
What is the number of water molecules present in a tiny drop of water (volume $0.0018 \mathrm{ml}$ ) at room temperature?
1 $4.84 \times 10^{17}$
2 $4.184 \times 10^{18}$
3 $6.023 \times 10^{19}$
4 $6.023 \times 10^{23}$
Explanation:
C Given data, Volume of water drop $=0.0018 \mathrm{~m} l$ Density of drop $=1000 \mathrm{~kg} / \mathrm{m}^{3}=1 \mathrm{~g} / \mathrm{m} l$ Mass of $0.0018 \mathrm{~m} l$ water $=$ Density $\times$ Volume $=1 \times 0.0018$ $=0.0018 \mathrm{~g}$ $\because$ We know that, Molar mass of water, $\mathrm{M}_{\mathrm{H}_{2} \mathrm{O}} =(2 \times 1)+16$ $=2+16$ $=18 \mathrm{~g} / \mathrm{mol}$ No. of moles is drop of water, $=\frac{\text { Mass }}{\text { Molar mass }}=\frac{0.0018}{18}$ $=0.0001 \mathrm{~mol}$ $\therefore$ No. of molecule present in drop of water $=6.023 \times 10^{23} \times 0.0001$ $\quad\left\{\because \mathrm{N}_{\mathrm{A}}=6.023 \times 10^{23} \text { molecule in } 1 \text { mole }\right\}$ $=6.023 \times 10^{19}$
NDA (II) 2008
Mechanical Properties of Fluids
143073
The radius of a soap bubble is increased from 1 $\mathrm{cm}$ to $4 \mathrm{~cm}$. If the surface tension of the soap solution is $0.021 \mathrm{Nm}^{-1}$, then the work done to increase the radius is
143077
If two soap bubbles of different radii are connected by a tube, then
1 Air flows from bigger bubble to the smaller bubble till sizes become equal
2 Air flows from bigger bubble to the smaller bubble till sizes are interchanged
3 Air flows from smaller bubble to bigger
4 There is no flow of air
Explanation:
C The additional pressure inside a soap bubble is inversely proportional to the radius of the soap bubble. $\mathrm{P} \propto \frac{1}{\mathrm{r}}$ $\mathrm{r}$ being the radius of bubble. If the pressure inside a smaller bubble is greater than the pressure inside a larger bubble. Thus if these two bubbles are connected by a tube, air will flow from the smaller bubble to the larger bubble and the larger bubble arouse at the Expense of the smaller bubble.
BCECE-2004
Mechanical Properties of Fluids
143019
A lead shot of $1 \mathrm{~mm}$ diameter falls through a long column of glycerine. The variation of its velocity $v$ with distance covered is represented by :
1
2
3 covered
4
Explanation:
A When lead shot of $1 \mathrm{~mm}$ diameter falls in glycerine through a long column, then due to gravity velocity of lead shot initially increases for some time. After some time due to viscosity of glycerine lead shot will attain a constant terminal velocity. Since, some upthrust force acting on lead shot due to glycerine so velocity will not be fully linear. So, the variation in velocity is given in figure.
143115
What is the number of water molecules present in a tiny drop of water (volume $0.0018 \mathrm{ml}$ ) at room temperature?
1 $4.84 \times 10^{17}$
2 $4.184 \times 10^{18}$
3 $6.023 \times 10^{19}$
4 $6.023 \times 10^{23}$
Explanation:
C Given data, Volume of water drop $=0.0018 \mathrm{~m} l$ Density of drop $=1000 \mathrm{~kg} / \mathrm{m}^{3}=1 \mathrm{~g} / \mathrm{m} l$ Mass of $0.0018 \mathrm{~m} l$ water $=$ Density $\times$ Volume $=1 \times 0.0018$ $=0.0018 \mathrm{~g}$ $\because$ We know that, Molar mass of water, $\mathrm{M}_{\mathrm{H}_{2} \mathrm{O}} =(2 \times 1)+16$ $=2+16$ $=18 \mathrm{~g} / \mathrm{mol}$ No. of moles is drop of water, $=\frac{\text { Mass }}{\text { Molar mass }}=\frac{0.0018}{18}$ $=0.0001 \mathrm{~mol}$ $\therefore$ No. of molecule present in drop of water $=6.023 \times 10^{23} \times 0.0001$ $\quad\left\{\because \mathrm{N}_{\mathrm{A}}=6.023 \times 10^{23} \text { molecule in } 1 \text { mole }\right\}$ $=6.023 \times 10^{19}$
NDA (II) 2008
Mechanical Properties of Fluids
143073
The radius of a soap bubble is increased from 1 $\mathrm{cm}$ to $4 \mathrm{~cm}$. If the surface tension of the soap solution is $0.021 \mathrm{Nm}^{-1}$, then the work done to increase the radius is
143077
If two soap bubbles of different radii are connected by a tube, then
1 Air flows from bigger bubble to the smaller bubble till sizes become equal
2 Air flows from bigger bubble to the smaller bubble till sizes are interchanged
3 Air flows from smaller bubble to bigger
4 There is no flow of air
Explanation:
C The additional pressure inside a soap bubble is inversely proportional to the radius of the soap bubble. $\mathrm{P} \propto \frac{1}{\mathrm{r}}$ $\mathrm{r}$ being the radius of bubble. If the pressure inside a smaller bubble is greater than the pressure inside a larger bubble. Thus if these two bubbles are connected by a tube, air will flow from the smaller bubble to the larger bubble and the larger bubble arouse at the Expense of the smaller bubble.
BCECE-2004
Mechanical Properties of Fluids
143019
A lead shot of $1 \mathrm{~mm}$ diameter falls through a long column of glycerine. The variation of its velocity $v$ with distance covered is represented by :
1
2
3 covered
4
Explanation:
A When lead shot of $1 \mathrm{~mm}$ diameter falls in glycerine through a long column, then due to gravity velocity of lead shot initially increases for some time. After some time due to viscosity of glycerine lead shot will attain a constant terminal velocity. Since, some upthrust force acting on lead shot due to glycerine so velocity will not be fully linear. So, the variation in velocity is given in figure.
