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In angiosperm, the haploid, diploid and triploid structures of a fertilized embryo sac sequentially are:
Option: 1
Synergids, Primary endosperm nucleus and zygote

Option: 2
Antipodals, synergids, and primary endosperm nucleus

Option: 3
Synergids, Zygote and Primary endosperm nucleus

Option: 4
Synergids, antipodals and Polar nuclei

The correct option is Option 3) Synergids, Zygote, and Primary endosperm nucleus.

In angiosperms, the fertilized embryo sac, which is the female gametophyte, contains various structures at different ploidy levels after fertilization. These structures include synergids, zygote, and the primary endosperm nucleus.

Synergids: Synergids are haploid cells found within the embryo sac. They are located at the micropylar end and play a role in guiding the pollen tube to the embryo sac during fertilization.

Zygote: After the pollen tube reaches the embryo sac and delivers the male gametes, fertilization occurs. The fusion of the sperm cell with the egg cell results in the formation of a zygote. The zygote is the first cell of the sporophytic generation and is diploid (2n) in ploidy.

Primary endosperm nucleus: The primary endosperm nucleus, also known as the central cell, is located at the center of the embryo sac. It contains two polar nuclei and is involved in the formation of endosperm. After fertilization, the primary endosperm nucleus undergoes triple fusion with one sperm cell, resulting in the formation of a triploid (3n) primary endosperm nucleus.

Therefore, the correct sequence of haploid, diploid, and triploid structures in a fertilized embryo sac is Synergids, Zygote, and Primary endosperm nucleus, as stated in Option 3.

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Posted by

Gunjita

A massless string connects two pulley of masses ' $2 \mathrm{~kg}$ ' and ' $1 \mathrm{~kg}$ ' respectively as shown in the figure.

The heavier pulley is fixed and free to rotate about its central axis while the other is free to rotate as well as translate. Find the acceleration of the lower pulley if the system was released from the rest. [Given, $g=10 \mathrm{~m} / \mathrm{s}^2$ ]
Option: 1
$\frac{4}{3} \mathrm{~gm} / \mathrm{s}^2$

Option: 2
$\frac{3}{2} \mathrm{~gm} / \mathrm{s}^2$

Option: 3
$\frac{3}{4} \mathrm{~gm} / \mathrm{s}^2$

Option: 4
$\frac{2}{3} \mathrm{~gm} / \mathrm{s}^2$

Not understanding sir

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Posted by

Raju vittal nandi

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The largest term in the expansion of $(3+2 x)^{50} \text {. where } x=\frac{1}{5} \text { is }$
Option: 1
$5^{\text {th }}$

Option: 2
$8^{\text {th }}$

Option: 3
$7^{\text {th }}$

Option: 4
$4^{\text {th }}$

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Posted by

In a common base amplifier circuit, calculate the change in base current if that in the emitter current is $\mathrm{2mA\: and\: \alpha =0.98}$

Option: 1
$\mathrm{0.04\: mA}$

Option: 2
$\mathrm{1.96\: mA}$

Option: 3
$\mathrm{0.98\: mA}$

Option: 4
$\mathrm{2\: mA}$

$\mathrm{ \alpha=\frac{\Delta I_c}{\Delta I_e} \Rightarrow \Delta I_c=0.98 \times 2=1.96 \mathrm{~mA} }$

$\mathrm{ \Delta I_b=\Delta I_e-\Delta I_c=2-1.96=0.04 \mathrm{~mA}}$

Hence option 1 is correct.

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Posted by

Ajit Kumar Dubey

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In the experiment, calculate the acceleration due to gravity, By using simple pendulum, which have time period of 0.3 second is measured from time of 50 oscillation with a watch of one second resolution. When length is 5 cm known to 1 mm accuracy find accuracy in the 'g'
Option: 1
15.33 %

Option: 2
12.33 %

Option: 3
13 %

Option: 4
15.55 %

$\begin{aligned} & T=2 \pi \sqrt{\frac{l}{g}} \\ & g=\frac{1}{4 \pi^2 l} \frac{T^2}{l}\Rightarrow \frac{\Delta g}{T}= \frac{2 \Delta T}{T}+\frac{\Delta l}{l} \\ & \frac{\Delta g}{g}=2 \times \frac{1}{50 \times 0.3}+\frac{1 \mathrm{~mm}}{5 \mathrm{~cm}} \\ & =0.1533 \\ & \frac{\Delta g}{g} \times 100=15.33 \% . \end{aligned}$

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Posted by

Kshitij

A sample of perfect gas is compressed isothermally to half its volume. If it is compressed adiabatically to the same volume, the final pressure of the gas will be
Option: 1
more

Option: 2
less

Option: 3
same

Option: 4
more or less depending on the initial temperature of the gas

For isothermal compression, $\mathrm{p_{2}=\frac{p V_{1}}{V_{2}}}$

For adiabatic compression, $\mathrm{p_{2}=p_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma}}$

Since, $\mathrm{\gamma=\frac{C_{p}}{C_{V}}>1}$ , hence $\mathrm{p_{2}}$ will be more as compared to that of isothermal compression.

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Posted by

Rakesh

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An ideal is taken through a cycle $\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C} \rightarrow \mathrm{A}$ as shown in Figure if the net heat supplied in the cycle is $5 \mathrm{~J}$ , then work done by the gas in the process $\mathrm{C} \rightarrow \mathrm{A}$ is

Option: 1
$-5 \mathrm{~J}$

Option: 2
$-10 \mathrm{~J}$

Option: 3
$-15 \mathrm{~J}$

Option: 4
$-20 \mathrm{~J}$

Work done $=$ area under the curve

$10 \mathrm{~J} ; 5=\mathrm{W}_{\mathrm{CA}}+10$ or $\mathrm{W}_{\mathrm{CA}}=-5 \mathrm{~J}$

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Posted by

Sayak

Two moles of ideal helium gas are in a rubber balloon at $30^{\circ} \mathrm{C}$ . The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to $35^{\circ} \mathrm{C}$ . The amount of heat required in raising the temperature is nearly (take R = 8.31 J/mol.K)
Option: 1
62J

Option: 2
104J

Option: 3
124J

Option: 4
208J

$\mathrm{\begin{aligned} & \Delta Q=n C_P \Delta T \\ & =2\left(\frac{3}{2} R+R\right) \Delta T \\ & =2\left[\frac{3}{2} R+R\right] \times 5 \\ & =2 \times \frac{5}{2} \times 8.31 \times 5 \\ & =208 \mathrm{~J} \end{aligned}}$