Physics 1 Marks Question

Let the external Edge of the cubical box be a m.

Total surface area of the cube = 6a?

The volume of the iron in the box = 6a² × 1 × 10^-2

= 6  × 10^-2 a²m²

Now, The mass of iron in the box = Volume x density

= 6 × 10^-2 × a² × 8000

= 48a²kg

Weight = 48 × 10 = 480N.

The volume of the water displaced = a²cm² [lt will be equal to volume of cube displaced].

$\therefore$ Upthrust = a² × 1000 × 10

= 10⁴a²

Now, 10⁴a² = 480

a = 48 × 10^-2m.

Therefore, a = 4.8cm.

Now, the lead piece does not displace the water, as per as our

mathematical calculations in last question. Only the wooden block

displaces.

Thus, the mass = 250cm² × 1g/cm²

= 250g

$\therefore$ Upthrust = 250 × g.

Now, (200 + a) × g = 250g

where, a is the mass of the metal lead.

a = 250 - 200

a = 50g

Let the mass of the lead piece be a kg.

Therefore, the total mass of the wood and the lead = 0.2 + akg. Density of lead = Density of water × specific gravity

= 1000 × 11.3 = 11300

Now, The volume of the lead piece $=\frac{\text{a}}{11300}\text{m}^3$

 When the wooden block is just allowed to float in water, it displaces a

volume of water equal to its volume =$\frac{200}{0.8}\text{cm}^3=250\text{cm}^3$

$= 250\text{cm}^2 = 0.25\text{m}^2 $

Now, The Total volume of the water displaced $=0.25+\frac{\text{a}}{11300}​​​​​​\text{cm}^2$

Upthrust $= \text{V}\rho\text{g}$

$=\Big(250+\frac{\text{a}}{11300}\Big)\times 1000 \times 10$

Now:

$(0.2+\text{a})\times 10 = \Big(250+\frac{\text{x}}{11300}\Big)\times 1000\times10$

$\therefore \text{a}=54.8\text{g}$$$

Let I be the minimum edge of the block.

Weight of the metal and ice = weight of the water displaced

(0.5 + 0.91%) = 1³ × 1 × g

(0.91 + 500) = 1³

$\rho=5000$

$\rho = 17.099\text{cm}$

Hence, minimum edge f the block shoud be 17.099cm.

Since the discharge through the tube remains the same i.e. 1cm/s, the speeds at A and B will be the same.

1. The speed at A,$\text{v}=\frac{1}{0.04}\text{cm}/\text{s}=\frac{100}{4}\text{cm}/\text{s}=25\text{cm}/\text{s}$ 

2. The speed at B, $\text{v}'=\frac{1}{0.02}\text{cm}/\text{s}=\frac{100}{2}\text{cm}/\text{s}=50\text{cm}/\text{s}$

3. Since the end B is upward, we take the end A as a reference level for the height. So the height of end A = 0 and the height of end B = h.

$=\frac{15}{16}\text{cm}$

$=\frac{15}{1600}\text{m}.$

From the Bernoulli's theorem

$\text{p}_\text{a}+\frac{1}{2}\rho\text{v}^2=\text{p}_\beta+\frac{1}{2}\rho\text{v}'^2+\rho\text{gh}$

$\Rightarrow\text{p}_\text{a}-\text{p}_\beta=\frac{1}{2}\text{p}\big(\text{v}^1-\text{v}^2\big)+\rho\text{gh}$

$=\frac{1}{2}\times1000\times\big(0.50^2-0.25^2\big)+1000\times10\times\frac{15}{1600}\text{N}/\text{m}^2$

$=\frac{1}{2}\times1000\times0.75\times0.25+\frac{1500}{16}\text{N}/\text{m}^2$

$=93.75+93.75\text{N}/\text{m}^2$

$=187.5\text{N}/\text{m}^2\approx188\text{N}/\text{m}^2$

1. Let the reference for height be the level of B. Height of A.

$=\text{h}=5\text{cm}=0.05\text{m}.\text{v}_\text{a}=10\text{cm}/\text{s}=0.10\text{cm}/\text{s}.$

Area of cross- section at $\text{A}=1\text{cm}^2=\frac{1}{10000}\text{m}^2.$

Hence the discharge $\text{Q}=0.10\times\frac{1}{10000}\text{m}^3=10^{-5}\text{m}^3$

Area of cross- section at $\text{B}=0.5\text{cm}^2=\frac{0.5}{10000}\text{m}^2$

The velocity at $\text{B}=\text{v}_\beta=\frac{10^{-5}}{\Big(\frac{0.5}{10000}\Big)}\text{m}/\text{s}=\frac{1}{5}\text{m}/\text{s}$

$=100\times\frac{1}{5}\text{cm}/\text{s}=20\text{cm}/\text{s}$

2. Let the pressures at A and B be Pa and PB. From the Bernoulli's theorem,

$\text{p}_\text{a}+\rho\text{gh}+\frac{1}{2}\rho\text{V}\text{a}^2=\text{p}_\beta+\frac{1}{2}\rho\beta^2+\frac{1}{2}\rho\text{v}\beta^2$

$\text{p}_\beta-\text{p}_\text{a}=\rho\text{gh}-\frac{1}{2}\rho\big(\text{V}\beta^2-\text{V}\text{a}^2\big)$

$=1000\times10\times0.05-\frac{1}{2}\times1000\times\big(0.2^2-0.1^2\big)\text{N}/\text{m}^2$

$=500-500\times(0.04-0.01)\text{N}/\text{m}^2$

$500-15\text{N}/\text{m}^2$

$=485\text{N}/\text{m}^2$

$=2\times\text{g}$

$=\pi\text{r}^2\times1\times\text{g}$

$\therefore$ Weight of the fluid displaced $\pi(20)^2\text{h}\times1\times\text{g}$

$2000\times\text{g}=\pi(10)^2\text{h}\times1\times\text{g}$

$2000=\pi(10)^2\text{h}$

$2000=\pi\times100\times\text{h}$

$\therefore\text{h}=\frac{20}{\pi}\text{cm}\cdot$

$\text{T}=2\pi\sqrt{\Big(\frac{\text{h}}{\text{g}}\Big)}$

$\text{T}=2\times\frac{22}{7}\times\sqrt{\Big(\frac{20}{\pi}\times980\Big)}$

$\therefore\text{T}=0.51$ seconds.

The spring balance measures the weight of an object. It will show the same weight

for the given iron and the wood block in the vacuum but in the air, a force of

buoyancy will act on both the blocks which will be equal to the weight of the air displaced.

The volume of iron block $=\frac{1}{7800}\text{m}^3$

The volume of wooden block $=\frac{1}{800}\text{m}^3$

The weight of air displaced by iron block$=\frac{1.293\times1}{7800}\text{Kg}$

$=\frac{1.293}{7800}\text{Kg}$

The weight of air displaced by wooden block $=\frac{1.293\times1}{800}\text{Kg}$

$=\frac{1.293}{800}\text{Kg}$

The apparent weight of the iron block shown in the spring balance $=\frac{1-1.293}{7800}\text{Kg}$

$=\frac{(7800-1.293)}{7800}\text{Kg}$

$=\frac{7798.707}{7800}\text{Kg}$

The apparent weight of the wooden block shown in the spring balance $=\frac{1-1.293}{800}\text{Kg}$

$=\frac{(800-1.293)}{800}\text{Kg}$

$=\frac{798.707}{800}\text{Kg}$

Hence the ratio of the apparent weights$=\frac{\Big(\frac{(7798.707)}{7800}\Big)}{\Big(\frac{798.707}{800}\Big)}$

$=\frac{\big(7898.707\times800\big)}{\big(7800\times798.707\big)}$

$=1.0015$