142820
A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and light slider supports a weight of $1.5 \times 10^{2} \mathrm{~N}$. The length of the slider is $30 \mathrm{~cm}$. What is the surface tension of the film?
1 $3 \times 10^{-3} \mathrm{Nm}^{-1}$
2 $2 \times 10^{-5} \mathrm{Nm}^{-1}$
3 $4 \times 10^{-4} \mathrm{Nm}^{-1}$
4 $2.5 \times 10^{-2} \mathrm{Nm}^{-1}$
Explanation:
D Given, $\mathrm{w}=1.5 \times 10^{2} \mathrm{~N}, l=30 \mathrm{~cm}$ We know that, Surface tension, $\mathrm{T}=\frac{\mathrm{F}}{l}$ Total length of the film Suspension, $(l)=2 \times 30 \mathrm{~cm}=0.60 \mathrm{~m}$ $\mathrm{F}=\mathrm{mg}$ $\mathrm{T}=\frac{\mathrm{mg}}{l}$ $\mathrm{T}=\frac{1.5 \times 10^{-2}}{0.6}$ $\mathrm{T}=2.5 \times 10^{-2} \mathrm{~N} / \mathrm{m}$ $\mathrm{T}=2.5 \times 10^{-2} \mathrm{Nm}^{-1}$
BCECE-2015
Mechanical Properties of Fluids
142821
The surface tension of soap solution is $\sigma$. What is the work done in blowing soap bubble of radius $r$ ?
1 $\pi r^{2} \sigma$
2 $2 \pi r^{2} \sigma$
3 $4 \pi r^{2} \sigma$
4 $8 \pi r^{2} \sigma$
Explanation:
D Work done $=$ Surface tension $\times 2(\Delta \mathrm{A})$ Change in Area $(\Delta \mathrm{A})=2 \times 4 \pi\left(\mathrm{R}_{1}^{2}-\mathrm{R}_{2}^{2}\right)$ $=2 \times 4 \pi \mathrm{r}^{2} \quad\left[\therefore \mathrm{R}_{1}^{2}-\mathrm{R}_{2}^{2}=\mathrm{r}^{2}\right]$ $\Delta \mathrm{A} \quad=8 \pi \mathrm{r}^{2}$ Work done $\quad=\sigma \times \Delta \mathrm{A}$ $=\sigma \times 8 \pi r^{2}$ $=8 \sigma \pi r^{2}$
BCECE-2009
Mechanical Properties of Fluids
142822
A soap film of surface tension $3 \times 10^{-2} \mathrm{Nm}^{-1}$ formed in rectangular frame, can support a straw. If the length of the film is $10 \mathrm{~cm}$, then the mass of the straw that film can support is-
142823
The surface tension of soap solution is $0.03 \mathrm{~N} / \mathrm{m}$. The amount of work done in forming a bubble of radius $5 \mathrm{~cm}$ is-
1 $3.77 \mathrm{~J}$
2 $1.885 \mathrm{~J}$
3 $0.95 \times 10^{-3} \mathrm{~J}$
4 $1.9 \times 10^{-3} \mathrm{~J}$
Explanation:
D Given that, $\mathrm{T}=0.03 \mathrm{~N} / \mathrm{m}$ $\mathrm{r}=5 \mathrm{~cm}=5 \times 10^{-2}$ We know that, Work done for bubble $(\mathrm{W})=\mathrm{T} \times 2(\mathrm{~A})$ $=\mathrm{T} \times 2\left(4 \pi \mathrm{r}^{2}\right)$ $=8 \pi \mathrm{r}^{2} \mathrm{~T}$ $\mathrm{W}=8 \times 3.14 \times\left(5 \times 10^{-2}\right)^{2} \times 0.03$ $\mathrm{W}=1.88 \times 10^{-3}=1.9 \times 10^{-3} \mathrm{~J}$
BCECE-2011
Mechanical Properties of Fluids
142824
The material of a wire has a density of 1.4 $\mathrm{g} / \mathrm{cm}^{3}$. It is not wetted by a liquid of surface tension $44 \mathrm{dyne} / \mathrm{cm}$, then the maximum radius of the wire which can float on the surface of liquid is-
1 $\frac{10}{28} \mathrm{~cm}$
2 $\frac{10}{14} \mathrm{~cm}$
3 $\frac{10}{7} \mathrm{~mm}$
4 $0.7 \mathrm{~cm}$
Explanation:
C We know that, Weight of the wire $=$ Force due to surface tension $\mathrm{mg}=\mathrm{T} \times(2 l)$ $(\rho \mathrm{V}) \mathrm{g}=2 \mathrm{~T} l$ $\rho\left(\pi \mathrm{r}^{2} l\right) \mathrm{g}=2 \mathrm{~T} l$ $\mathrm{r}=\sqrt{\left(\frac{2 \mathrm{~T}}{\pi \rho g}\right)}$ $= \sqrt{\frac{2 \times 44 \times 7}{22 \times 1.4 \times 980}}$ $= \frac{1}{7} \mathrm{~cm}$ $= \frac{10}{7} \mathrm{~mm}$
142820
A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and light slider supports a weight of $1.5 \times 10^{2} \mathrm{~N}$. The length of the slider is $30 \mathrm{~cm}$. What is the surface tension of the film?
1 $3 \times 10^{-3} \mathrm{Nm}^{-1}$
2 $2 \times 10^{-5} \mathrm{Nm}^{-1}$
3 $4 \times 10^{-4} \mathrm{Nm}^{-1}$
4 $2.5 \times 10^{-2} \mathrm{Nm}^{-1}$
Explanation:
D Given, $\mathrm{w}=1.5 \times 10^{2} \mathrm{~N}, l=30 \mathrm{~cm}$ We know that, Surface tension, $\mathrm{T}=\frac{\mathrm{F}}{l}$ Total length of the film Suspension, $(l)=2 \times 30 \mathrm{~cm}=0.60 \mathrm{~m}$ $\mathrm{F}=\mathrm{mg}$ $\mathrm{T}=\frac{\mathrm{mg}}{l}$ $\mathrm{T}=\frac{1.5 \times 10^{-2}}{0.6}$ $\mathrm{T}=2.5 \times 10^{-2} \mathrm{~N} / \mathrm{m}$ $\mathrm{T}=2.5 \times 10^{-2} \mathrm{Nm}^{-1}$
BCECE-2015
Mechanical Properties of Fluids
142821
The surface tension of soap solution is $\sigma$. What is the work done in blowing soap bubble of radius $r$ ?
1 $\pi r^{2} \sigma$
2 $2 \pi r^{2} \sigma$
3 $4 \pi r^{2} \sigma$
4 $8 \pi r^{2} \sigma$
Explanation:
D Work done $=$ Surface tension $\times 2(\Delta \mathrm{A})$ Change in Area $(\Delta \mathrm{A})=2 \times 4 \pi\left(\mathrm{R}_{1}^{2}-\mathrm{R}_{2}^{2}\right)$ $=2 \times 4 \pi \mathrm{r}^{2} \quad\left[\therefore \mathrm{R}_{1}^{2}-\mathrm{R}_{2}^{2}=\mathrm{r}^{2}\right]$ $\Delta \mathrm{A} \quad=8 \pi \mathrm{r}^{2}$ Work done $\quad=\sigma \times \Delta \mathrm{A}$ $=\sigma \times 8 \pi r^{2}$ $=8 \sigma \pi r^{2}$
BCECE-2009
Mechanical Properties of Fluids
142822
A soap film of surface tension $3 \times 10^{-2} \mathrm{Nm}^{-1}$ formed in rectangular frame, can support a straw. If the length of the film is $10 \mathrm{~cm}$, then the mass of the straw that film can support is-
142823
The surface tension of soap solution is $0.03 \mathrm{~N} / \mathrm{m}$. The amount of work done in forming a bubble of radius $5 \mathrm{~cm}$ is-
1 $3.77 \mathrm{~J}$
2 $1.885 \mathrm{~J}$
3 $0.95 \times 10^{-3} \mathrm{~J}$
4 $1.9 \times 10^{-3} \mathrm{~J}$
Explanation:
D Given that, $\mathrm{T}=0.03 \mathrm{~N} / \mathrm{m}$ $\mathrm{r}=5 \mathrm{~cm}=5 \times 10^{-2}$ We know that, Work done for bubble $(\mathrm{W})=\mathrm{T} \times 2(\mathrm{~A})$ $=\mathrm{T} \times 2\left(4 \pi \mathrm{r}^{2}\right)$ $=8 \pi \mathrm{r}^{2} \mathrm{~T}$ $\mathrm{W}=8 \times 3.14 \times\left(5 \times 10^{-2}\right)^{2} \times 0.03$ $\mathrm{W}=1.88 \times 10^{-3}=1.9 \times 10^{-3} \mathrm{~J}$
BCECE-2011
Mechanical Properties of Fluids
142824
The material of a wire has a density of 1.4 $\mathrm{g} / \mathrm{cm}^{3}$. It is not wetted by a liquid of surface tension $44 \mathrm{dyne} / \mathrm{cm}$, then the maximum radius of the wire which can float on the surface of liquid is-
1 $\frac{10}{28} \mathrm{~cm}$
2 $\frac{10}{14} \mathrm{~cm}$
3 $\frac{10}{7} \mathrm{~mm}$
4 $0.7 \mathrm{~cm}$
Explanation:
C We know that, Weight of the wire $=$ Force due to surface tension $\mathrm{mg}=\mathrm{T} \times(2 l)$ $(\rho \mathrm{V}) \mathrm{g}=2 \mathrm{~T} l$ $\rho\left(\pi \mathrm{r}^{2} l\right) \mathrm{g}=2 \mathrm{~T} l$ $\mathrm{r}=\sqrt{\left(\frac{2 \mathrm{~T}}{\pi \rho g}\right)}$ $= \sqrt{\frac{2 \times 44 \times 7}{22 \times 1.4 \times 980}}$ $= \frac{1}{7} \mathrm{~cm}$ $= \frac{10}{7} \mathrm{~mm}$
142820
A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and light slider supports a weight of $1.5 \times 10^{2} \mathrm{~N}$. The length of the slider is $30 \mathrm{~cm}$. What is the surface tension of the film?
1 $3 \times 10^{-3} \mathrm{Nm}^{-1}$
2 $2 \times 10^{-5} \mathrm{Nm}^{-1}$
3 $4 \times 10^{-4} \mathrm{Nm}^{-1}$
4 $2.5 \times 10^{-2} \mathrm{Nm}^{-1}$
Explanation:
D Given, $\mathrm{w}=1.5 \times 10^{2} \mathrm{~N}, l=30 \mathrm{~cm}$ We know that, Surface tension, $\mathrm{T}=\frac{\mathrm{F}}{l}$ Total length of the film Suspension, $(l)=2 \times 30 \mathrm{~cm}=0.60 \mathrm{~m}$ $\mathrm{F}=\mathrm{mg}$ $\mathrm{T}=\frac{\mathrm{mg}}{l}$ $\mathrm{T}=\frac{1.5 \times 10^{-2}}{0.6}$ $\mathrm{T}=2.5 \times 10^{-2} \mathrm{~N} / \mathrm{m}$ $\mathrm{T}=2.5 \times 10^{-2} \mathrm{Nm}^{-1}$
BCECE-2015
Mechanical Properties of Fluids
142821
The surface tension of soap solution is $\sigma$. What is the work done in blowing soap bubble of radius $r$ ?
1 $\pi r^{2} \sigma$
2 $2 \pi r^{2} \sigma$
3 $4 \pi r^{2} \sigma$
4 $8 \pi r^{2} \sigma$
Explanation:
D Work done $=$ Surface tension $\times 2(\Delta \mathrm{A})$ Change in Area $(\Delta \mathrm{A})=2 \times 4 \pi\left(\mathrm{R}_{1}^{2}-\mathrm{R}_{2}^{2}\right)$ $=2 \times 4 \pi \mathrm{r}^{2} \quad\left[\therefore \mathrm{R}_{1}^{2}-\mathrm{R}_{2}^{2}=\mathrm{r}^{2}\right]$ $\Delta \mathrm{A} \quad=8 \pi \mathrm{r}^{2}$ Work done $\quad=\sigma \times \Delta \mathrm{A}$ $=\sigma \times 8 \pi r^{2}$ $=8 \sigma \pi r^{2}$
BCECE-2009
Mechanical Properties of Fluids
142822
A soap film of surface tension $3 \times 10^{-2} \mathrm{Nm}^{-1}$ formed in rectangular frame, can support a straw. If the length of the film is $10 \mathrm{~cm}$, then the mass of the straw that film can support is-
142823
The surface tension of soap solution is $0.03 \mathrm{~N} / \mathrm{m}$. The amount of work done in forming a bubble of radius $5 \mathrm{~cm}$ is-
1 $3.77 \mathrm{~J}$
2 $1.885 \mathrm{~J}$
3 $0.95 \times 10^{-3} \mathrm{~J}$
4 $1.9 \times 10^{-3} \mathrm{~J}$
Explanation:
D Given that, $\mathrm{T}=0.03 \mathrm{~N} / \mathrm{m}$ $\mathrm{r}=5 \mathrm{~cm}=5 \times 10^{-2}$ We know that, Work done for bubble $(\mathrm{W})=\mathrm{T} \times 2(\mathrm{~A})$ $=\mathrm{T} \times 2\left(4 \pi \mathrm{r}^{2}\right)$ $=8 \pi \mathrm{r}^{2} \mathrm{~T}$ $\mathrm{W}=8 \times 3.14 \times\left(5 \times 10^{-2}\right)^{2} \times 0.03$ $\mathrm{W}=1.88 \times 10^{-3}=1.9 \times 10^{-3} \mathrm{~J}$
BCECE-2011
Mechanical Properties of Fluids
142824
The material of a wire has a density of 1.4 $\mathrm{g} / \mathrm{cm}^{3}$. It is not wetted by a liquid of surface tension $44 \mathrm{dyne} / \mathrm{cm}$, then the maximum radius of the wire which can float on the surface of liquid is-
1 $\frac{10}{28} \mathrm{~cm}$
2 $\frac{10}{14} \mathrm{~cm}$
3 $\frac{10}{7} \mathrm{~mm}$
4 $0.7 \mathrm{~cm}$
Explanation:
C We know that, Weight of the wire $=$ Force due to surface tension $\mathrm{mg}=\mathrm{T} \times(2 l)$ $(\rho \mathrm{V}) \mathrm{g}=2 \mathrm{~T} l$ $\rho\left(\pi \mathrm{r}^{2} l\right) \mathrm{g}=2 \mathrm{~T} l$ $\mathrm{r}=\sqrt{\left(\frac{2 \mathrm{~T}}{\pi \rho g}\right)}$ $= \sqrt{\frac{2 \times 44 \times 7}{22 \times 1.4 \times 980}}$ $= \frac{1}{7} \mathrm{~cm}$ $= \frac{10}{7} \mathrm{~mm}$
142820
A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and light slider supports a weight of $1.5 \times 10^{2} \mathrm{~N}$. The length of the slider is $30 \mathrm{~cm}$. What is the surface tension of the film?
1 $3 \times 10^{-3} \mathrm{Nm}^{-1}$
2 $2 \times 10^{-5} \mathrm{Nm}^{-1}$
3 $4 \times 10^{-4} \mathrm{Nm}^{-1}$
4 $2.5 \times 10^{-2} \mathrm{Nm}^{-1}$
Explanation:
D Given, $\mathrm{w}=1.5 \times 10^{2} \mathrm{~N}, l=30 \mathrm{~cm}$ We know that, Surface tension, $\mathrm{T}=\frac{\mathrm{F}}{l}$ Total length of the film Suspension, $(l)=2 \times 30 \mathrm{~cm}=0.60 \mathrm{~m}$ $\mathrm{F}=\mathrm{mg}$ $\mathrm{T}=\frac{\mathrm{mg}}{l}$ $\mathrm{T}=\frac{1.5 \times 10^{-2}}{0.6}$ $\mathrm{T}=2.5 \times 10^{-2} \mathrm{~N} / \mathrm{m}$ $\mathrm{T}=2.5 \times 10^{-2} \mathrm{Nm}^{-1}$
BCECE-2015
Mechanical Properties of Fluids
142821
The surface tension of soap solution is $\sigma$. What is the work done in blowing soap bubble of radius $r$ ?
1 $\pi r^{2} \sigma$
2 $2 \pi r^{2} \sigma$
3 $4 \pi r^{2} \sigma$
4 $8 \pi r^{2} \sigma$
Explanation:
D Work done $=$ Surface tension $\times 2(\Delta \mathrm{A})$ Change in Area $(\Delta \mathrm{A})=2 \times 4 \pi\left(\mathrm{R}_{1}^{2}-\mathrm{R}_{2}^{2}\right)$ $=2 \times 4 \pi \mathrm{r}^{2} \quad\left[\therefore \mathrm{R}_{1}^{2}-\mathrm{R}_{2}^{2}=\mathrm{r}^{2}\right]$ $\Delta \mathrm{A} \quad=8 \pi \mathrm{r}^{2}$ Work done $\quad=\sigma \times \Delta \mathrm{A}$ $=\sigma \times 8 \pi r^{2}$ $=8 \sigma \pi r^{2}$
BCECE-2009
Mechanical Properties of Fluids
142822
A soap film of surface tension $3 \times 10^{-2} \mathrm{Nm}^{-1}$ formed in rectangular frame, can support a straw. If the length of the film is $10 \mathrm{~cm}$, then the mass of the straw that film can support is-
142823
The surface tension of soap solution is $0.03 \mathrm{~N} / \mathrm{m}$. The amount of work done in forming a bubble of radius $5 \mathrm{~cm}$ is-
1 $3.77 \mathrm{~J}$
2 $1.885 \mathrm{~J}$
3 $0.95 \times 10^{-3} \mathrm{~J}$
4 $1.9 \times 10^{-3} \mathrm{~J}$
Explanation:
D Given that, $\mathrm{T}=0.03 \mathrm{~N} / \mathrm{m}$ $\mathrm{r}=5 \mathrm{~cm}=5 \times 10^{-2}$ We know that, Work done for bubble $(\mathrm{W})=\mathrm{T} \times 2(\mathrm{~A})$ $=\mathrm{T} \times 2\left(4 \pi \mathrm{r}^{2}\right)$ $=8 \pi \mathrm{r}^{2} \mathrm{~T}$ $\mathrm{W}=8 \times 3.14 \times\left(5 \times 10^{-2}\right)^{2} \times 0.03$ $\mathrm{W}=1.88 \times 10^{-3}=1.9 \times 10^{-3} \mathrm{~J}$
BCECE-2011
Mechanical Properties of Fluids
142824
The material of a wire has a density of 1.4 $\mathrm{g} / \mathrm{cm}^{3}$. It is not wetted by a liquid of surface tension $44 \mathrm{dyne} / \mathrm{cm}$, then the maximum radius of the wire which can float on the surface of liquid is-
1 $\frac{10}{28} \mathrm{~cm}$
2 $\frac{10}{14} \mathrm{~cm}$
3 $\frac{10}{7} \mathrm{~mm}$
4 $0.7 \mathrm{~cm}$
Explanation:
C We know that, Weight of the wire $=$ Force due to surface tension $\mathrm{mg}=\mathrm{T} \times(2 l)$ $(\rho \mathrm{V}) \mathrm{g}=2 \mathrm{~T} l$ $\rho\left(\pi \mathrm{r}^{2} l\right) \mathrm{g}=2 \mathrm{~T} l$ $\mathrm{r}=\sqrt{\left(\frac{2 \mathrm{~T}}{\pi \rho g}\right)}$ $= \sqrt{\frac{2 \times 44 \times 7}{22 \times 1.4 \times 980}}$ $= \frac{1}{7} \mathrm{~cm}$ $= \frac{10}{7} \mathrm{~mm}$
142820
A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and light slider supports a weight of $1.5 \times 10^{2} \mathrm{~N}$. The length of the slider is $30 \mathrm{~cm}$. What is the surface tension of the film?
1 $3 \times 10^{-3} \mathrm{Nm}^{-1}$
2 $2 \times 10^{-5} \mathrm{Nm}^{-1}$
3 $4 \times 10^{-4} \mathrm{Nm}^{-1}$
4 $2.5 \times 10^{-2} \mathrm{Nm}^{-1}$
Explanation:
D Given, $\mathrm{w}=1.5 \times 10^{2} \mathrm{~N}, l=30 \mathrm{~cm}$ We know that, Surface tension, $\mathrm{T}=\frac{\mathrm{F}}{l}$ Total length of the film Suspension, $(l)=2 \times 30 \mathrm{~cm}=0.60 \mathrm{~m}$ $\mathrm{F}=\mathrm{mg}$ $\mathrm{T}=\frac{\mathrm{mg}}{l}$ $\mathrm{T}=\frac{1.5 \times 10^{-2}}{0.6}$ $\mathrm{T}=2.5 \times 10^{-2} \mathrm{~N} / \mathrm{m}$ $\mathrm{T}=2.5 \times 10^{-2} \mathrm{Nm}^{-1}$
BCECE-2015
Mechanical Properties of Fluids
142821
The surface tension of soap solution is $\sigma$. What is the work done in blowing soap bubble of radius $r$ ?
1 $\pi r^{2} \sigma$
2 $2 \pi r^{2} \sigma$
3 $4 \pi r^{2} \sigma$
4 $8 \pi r^{2} \sigma$
Explanation:
D Work done $=$ Surface tension $\times 2(\Delta \mathrm{A})$ Change in Area $(\Delta \mathrm{A})=2 \times 4 \pi\left(\mathrm{R}_{1}^{2}-\mathrm{R}_{2}^{2}\right)$ $=2 \times 4 \pi \mathrm{r}^{2} \quad\left[\therefore \mathrm{R}_{1}^{2}-\mathrm{R}_{2}^{2}=\mathrm{r}^{2}\right]$ $\Delta \mathrm{A} \quad=8 \pi \mathrm{r}^{2}$ Work done $\quad=\sigma \times \Delta \mathrm{A}$ $=\sigma \times 8 \pi r^{2}$ $=8 \sigma \pi r^{2}$
BCECE-2009
Mechanical Properties of Fluids
142822
A soap film of surface tension $3 \times 10^{-2} \mathrm{Nm}^{-1}$ formed in rectangular frame, can support a straw. If the length of the film is $10 \mathrm{~cm}$, then the mass of the straw that film can support is-
142823
The surface tension of soap solution is $0.03 \mathrm{~N} / \mathrm{m}$. The amount of work done in forming a bubble of radius $5 \mathrm{~cm}$ is-
1 $3.77 \mathrm{~J}$
2 $1.885 \mathrm{~J}$
3 $0.95 \times 10^{-3} \mathrm{~J}$
4 $1.9 \times 10^{-3} \mathrm{~J}$
Explanation:
D Given that, $\mathrm{T}=0.03 \mathrm{~N} / \mathrm{m}$ $\mathrm{r}=5 \mathrm{~cm}=5 \times 10^{-2}$ We know that, Work done for bubble $(\mathrm{W})=\mathrm{T} \times 2(\mathrm{~A})$ $=\mathrm{T} \times 2\left(4 \pi \mathrm{r}^{2}\right)$ $=8 \pi \mathrm{r}^{2} \mathrm{~T}$ $\mathrm{W}=8 \times 3.14 \times\left(5 \times 10^{-2}\right)^{2} \times 0.03$ $\mathrm{W}=1.88 \times 10^{-3}=1.9 \times 10^{-3} \mathrm{~J}$
BCECE-2011
Mechanical Properties of Fluids
142824
The material of a wire has a density of 1.4 $\mathrm{g} / \mathrm{cm}^{3}$. It is not wetted by a liquid of surface tension $44 \mathrm{dyne} / \mathrm{cm}$, then the maximum radius of the wire which can float on the surface of liquid is-
1 $\frac{10}{28} \mathrm{~cm}$
2 $\frac{10}{14} \mathrm{~cm}$
3 $\frac{10}{7} \mathrm{~mm}$
4 $0.7 \mathrm{~cm}$
Explanation:
C We know that, Weight of the wire $=$ Force due to surface tension $\mathrm{mg}=\mathrm{T} \times(2 l)$ $(\rho \mathrm{V}) \mathrm{g}=2 \mathrm{~T} l$ $\rho\left(\pi \mathrm{r}^{2} l\right) \mathrm{g}=2 \mathrm{~T} l$ $\mathrm{r}=\sqrt{\left(\frac{2 \mathrm{~T}}{\pi \rho g}\right)}$ $= \sqrt{\frac{2 \times 44 \times 7}{22 \times 1.4 \times 980}}$ $= \frac{1}{7} \mathrm{~cm}$ $= \frac{10}{7} \mathrm{~mm}$
